What is the slope of 5y =- 10?

Answer:

Step-by-step explanation:

Evaluate 5 - 3(a3 – b2)2 when a = 3 and b = 5.

5 - 3 × (a³-b²)²                           a=3 and b=5

5 - 3 × (3³ - 5²)² =

5 - 3 × (27  - 25)² =

5 - 3 × (2)² =

5 - 3 × 4 =

5 - 12 =

-7

158/4 as a mixed number in the simplest term is  [tex]39\frac{3}{4}[/tex]

The given fraction is 158/4

The number of times that 158 can be divided by 4 = 39

4 x 39 = 156

The remainder = 159 - 156

The remainder = 3

The mixed number is of the form:

[tex]Quotient\frac{Remaider}{Divisor}[/tex]

Therefore, the mixed number is [tex]39\frac{3}{4}[/tex]

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The distance between the two stations is 360km.

Speed is the rate of change of distance.

Rate is a measure of one quantity against another in this case distance and time.

Analysis:

time at station A = 12:24

time at station B = 14:12

time spent = 14:12 - 12:24 = 1 hour 48 minutes = convert 48 minutes to hour

we divide 48 by 60 = 0.8

Total time = 1.8 hours

Distance = speed x time = 200 x 1.8 = 360 km

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Answer:

d. x =0 or x = -6

Step-by-step explanation:

x(x + 6) = 0

This is telling us that either x is 0 or x is -6, because:

1) when x = 0, 0*(0+6)=0, and

2) when x = -6, -6*(-6+6) = 0;  -6(0) = 0

Considering the critical points of the function, we have that:

The critical points of a function are the values of x for which:

[tex]f^{\prime}(x) = 0[/tex]

In this problem, the function is:

[tex]f(x) = 3x^3 - 16x + 2[/tex]

The derivative is:

[tex]f^{\prime}{x} = 6x^2 - 16[/tex]

The critical points are given as follows:

[tex]6x^2 - 16 = 0[/tex]

[tex]x^2 = \frac{16}{6}[/tex]

[tex]x = \pm \sqrt{\frac{16}{6}}[/tex]

[tex]x = \pm 1.63[/tex]

For x < -1.63, one example of the derivative is:

[tex]f^{\prime}{-2} = 6(-2)^2 - 16 = 8[/tex]

Positive, hence increasing.

For -1.63 < x < 1.63, one example of the derivative is:

[tex]f^{\prime}{0} = 6(0)^2 - 16 = -16[/tex]

Negative, hence decreasing.

For x > 1.63, one example of the derivative is:

[tex]f^{\prime}{2} = 6(2)^2 - 16 = 8[/tex]

Positive, hence increasing.

Hence:

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The value of a is -3 and the value of k is 10

A parabola is a U-shaped plane curve where any point is equidistant from a fixed point (known as the focus) and from a fixed line known as the directrix. The parabola is an integral part of the conic section topic

The section of a right circular cone by a plane parallel to the generator of the cone is a parabola. It is the location of a point that moves so that the distance from the fixed point (the focus) is equal to the distance from the fixed line (the directrix).

The fixed point is called the focus

A fixed line is called a directrix

Solving an algebraic equation is the process of finding a number or set of numbers that, when substituted for the variables in the equation, reduce them to an identity. Such a number is called the root of the equation

Here the equation y = a(x -2)^2 + k...........(1)

and two points A(1,7) and B(4, -2) is given

We need to find the value of a and k

Put A(1,7) in equation (1) , we get,

7 = a + k ,therefore k = 7 - a......(2)

Put B(4, -2) in equation (1), we get,

-2 = 4a + k........(3)

Put k= 7-a in (3)

-2 = 4a+7 -a

3a = -9

a= -3

From equation (2) , we get k = 7-(-3)= 10

Hence the value of k is 10 and the value of a is -3

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For the given points to lie on the parabola,

a = -3 and k = 10.

An equation of a curve that has a point on it that is equally spaced from a fixed point and a fixed line is referred to as a parabola. The parabola's fixed line and fixed point are together referred to as the directrix and focus, respectively. It's also crucial to remember that the fixed point is not located on the fixed line. A parabola is a locus of any point that is equally distant from a given point (focus) and a certain line (directrix).

According to the question,

Equation of parabola : y = a[tex](x-2)^{2}[/tex] + k

Points A(1,7) and B(4,-2)

For the points to lie on the parabola,

7 = a[tex](1-2)^{2}[/tex]+k

7 = a + k

Similarly,

-2 = a[tex](4-2)^{2}[/tex] + k

-2 = 4a + k

On solving the two equations simultaneously, we get,

a = -3

k = 10

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The equation can be used to find other combinations of x and y is  y^2 = (1/16)x^3

The direct variation from the square of y to the cube of x is represented as:

y^2 = kx^3

Where k represents the variation constant.

When x = 4, y = 2.

So, we have:

2^2 = k * 4^3

This gives

4 = 64k

Divide both sides by 64

k = 1/16

Substitute k = 1/16 in y^2 = kx^3

y^2 = (1/16)x^3

Hence, the equation can be used to find other combinations of x and y is  y^2 = (1/16)x^3

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Answer:

There are 24 frames of animation in 1 second of anime.    Step-by-step explanation:    There are 60 seconds in 1 minute, and 60 minutes in 1 hour. 60×60=3600, so there are 3600 seconds in 1 hour.Since 3600 seconds and 1 hour is the same, there are 86400 frames of animation in 3600 seconds of anime. To find out how many frames are in one second of anime, divide 86400 by 3600.86400 ÷ 3600 = 24There are 24 frames of animation in 1 second of anime.

Taking a quotient, we will see that she can make 7 bowls of cereal (and some leftover milk).

We know that for each bowl, she needs 1/2 cups of milk.

And we also know that she has a total of (3 + 3/5) cups of milk.

To know how many bowls she can make, we need to take the quotient between the total that she has and the amount that she needs for each bowl:

(3 + 3/5)/(1/2)

We can rewrite the total as:

3 + 3/5 = 15/5 + 3/5 = 18/5

Then the quotient becomes:

(18/5)/(1/2) = (18/5)*2 = 36/5 = 35/5 + 1/5 = 7 + 1/5

So she can make 7 bowls of cereal (and some leftover milk).

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The diameter of circle O in the image given is: AB.

The diameter of a circle can be referred to as the largest chord in a circle which is the line segment that passes through the center of a circle with both ends on the circle.

In the image given, AB is the largest chord and also passes through the center of the circle, O.

Thus, the diameter is: AB.

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Using an exponential function, it is found that Mary's initial investment was of $3,030.

An exponential function is modeled by:

[tex]y = ab^x[/tex].

In which:

Her investment model, in thousands of dollars, is:

[tex]M(x) = 3.03(1.28)^{2x}[/tex]

Then a = 3.03, since we measure the amount in thousands of dollars, Mary's initial investment was of $3,030.

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The value of the angle will be C. 20°

From the information given, it was stated that the protractor showed an angle going through the tenth tick mark after ten degrees.

This means the value of the angle will be:

= 10° + 10°

= 20°

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The experimental probability that the subsequent booth competitor will receive a reward is 3/8.

The possibility of an event occurring is determined by probability.

The probability of the occurrence ranges from 0 to 1.

If the event doesn't happen, it = 0;

otherwise, it = 1.

For instance, the likelihood that it will storm on Sunday ranges from 0 to 1. If it storms, the event is given a value of 1. If it doesn't, the event is given a value of zero.

Depending on the outcome of a study that has been run several times, the experimental probability is calculated.

The number of winners in the games divided by the total number of competitors in those games determines the probability that the following competitor will earn a reward.

So, here the probability = 6/16 = 3/8 (by dividing both the numerator and the denominator by 2).

Therefore, the final solution is P= 3/8 .

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Answer:

Step-by-step explanation:

The value of t based on the information about the sphere is 1.3π.

It should be noted that the volume of a sphere is 4/3πr³. In this case, it's divided into 8 equal parts.

Volume of each part = 1/8 × 4/3πr³

= 1/8 × 4/3 × π × 8

= 4/3π

= 1.3πin³

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Answer:

The longest side of room is 24m

The statement that correctly describes the horizontal asymptote of g(x) is:

Limit of g (x) as x approaches plus-or-minus infinity = 6, so g(x) has an asymptote at y = 6.

In this problem, the function is:

[tex]g(x) = \frac{42x^3 - 15}{7x^3 - 4x^2 - 3}[/tex]

The horizontal asymptote is given as follows:

[tex]y = \lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} \frac{42x^3 - 15}{7x^3 - 4x^2 - 3} = \lim_{x \rightarrow \infty} \frac{42x^3}{7x^3} = \lim_{x \rightarrow \infty} 6 = 6[/tex]

Hence the correct statement is:

Limit of g (x) as x approaches plus-or-minus infinity = 6, so g(x) has an asymptote at y = 6.

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1) Difference of  the populations  of Germany and  Brazil = 1.16 X 10^8

2) Difference of  the populations  of Brazil and the  United States = 1.2 X 10^8

3) Sum of the  populations of  Brazil and  Germany = 2.76 X 10^8

4) Sum of the  populations of  the United States  and Brazil = 5.12 X 10^8

According to the statement

we have given that the populations of  three countries which are

United States is 3.16 × 10^8 and  Brazil is 1.96 × 10^8 and Germany is 8.0 × 10^7

And we have to arrange the sums and differences in increasing order of their values.

so, we have to find the

a) the difference of  the populations  of Germany and  Brazil

So,

Difference of  the populations  of Germany and  Brazil =  1.96 X 10^8 - 8 X 10^7 = 116000000 = 1.16 X 10^8

And then we have to find the

b) the sum of the  populations of  the United States  and Brazil

So,

Sum of the  populations of  the United States  and Brazil = 3.16 X 10^8 + 1.96 X 10^8 = 512000000 = 5.12 X 10^8

And then we have to find the

c) the difference of  the populations  of Brazil and the  United States

So,

Difference of  the populations  of Brazil and the  United States = 3.16 X 10^8 - 1.96 X 10^8 = 120000000 = 1.2 X 10^8

And then we have to find the

d) the sum of the  populations of  Brazil and  Germany

So,

Sum of the  populations of  Brazil and  Germany:  1.96 X 10^8 + 8 X 10^7 = 276000000 = 2.76 X 10^8

NOW, we arrange the sum and difference in increasing order of their values.

So,

1) Difference of  the populations  of Germany and  Brazil = 1.16 X 10^8

2) Difference of  the populations  of Brazil and the  United States = 1.2 X 10^8

3) Sum of the  populations of  Brazil and  Germany = 2.76 X 10^8

4) Sum of the  populations of  the United States  and Brazil = 5.12 X 10^8

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i. The number of people who like banana only is 20.

ii. The number of people who like none of the fruits is zero.

iii. The number of people who like at most one fruit is 30.

The question has to do with sets.

A set a collection of well ordered items

Since we have 55 people, 3x + 5 people like banana and x + 5 people like apple. If all the people who like apple also like banana and if 2x + 15 people like at least one of fruit then, we have that

3x + 5 + x + 5 + 2x + 15 = 55

3x + 2x + x + 5 + 5 + 15 = 55

6x + 30 = 55

6x = 55 - 25

6x = 30

x = 30/6

x = 5

Since the number of people who like banana only is 3x + 5.

So, number of people who like banana only is n = 3x + 5

= 3(5) + 5

= 15 + 5

= 20

So, the number of people who like banana only is 20.

Since from the question, a person likes at least one fruit, either banana, apple or both. No one likes none of the fruits.

So, the number of people who like none of the fruits is zero.

To like at most one fruit, the person either likes apple or banana.

So, their sum is n = 3x + 5 + x + 5

= 4x + 10.

Since x = 5, we have n = 4x + 10.

= 4(5) + 10

= 20 + 10

= 30

So, the number of people who like at most one fruit is 30.

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The probablity that the sample's mean length is greate than 6.3 inches is0.8446.

Given mean of 6.5 inches,standard deviation of 0.5 inches and sample size of 46.

We have to calculate the probability that the sample's mean length is greater than 6.3 inches is 0.8446.

Probability is the likeliness of happening an event. It lies between 0 and 1.

Probability is the number of items divided by the total number of items.

We have to use z statistic in this question because the sample size is greater than 30.

μ=6.5

σ=0.5

n=46

z=X-μ/σ

where μ is mean and

σ is standard deviation.

First we have to find the p value from 6.3 to 6.5 and then we have to add 0.5 to it to find the required probability.

z=6.3-6.5/0.5

=-0.2/0.5

=-0.4

p value from z table is 0.3446

Probability that the mean length is greater than 6.3inches is 0.3446+0.5=0.8446.

Hence the probability that the mean length is greater than 6.3 inches is 0.8446.

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The error in the steps is that In step 1, she should have also distributed -3/8  over –16d, to get 3 + 6 d + 2 d = 24

Given the following equation as shown below

-3/8(-8-16d)+2d = 24

Step 1 Distribute -3/8 over the expression in parentheses

3 + 6d + 2d = 24

Simply the like terms

3 + 8d = 24

Subtract 3 from both sides of the equation.

8d = 24 - 3

8d = 21

d = 21/8

Hence the error in the steps is that In step 1, she should have also distributed -3/8  over –16d, to get 3 + 6 d + 2 d = 24

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The initial amount of the element is 16

The function is given as:

f(t) = 16(1/2)^t

Set t = 0, to determine the initial amount

f(0) = 16(1/2)^0

This gives

f(0) = 16 * 1

Evaluate

f(0) = 16

Hence, the initial amount of the element is 16

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[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

[tex]\qquad❖ \: \sf \:x = - \dfrac{ 9}{20} [/tex]

[tex]\textsf{ \underline{\underline{Steps to solve the problem} }:}[/tex]

[tex]\qquad❖ \: \sf \:2(3x + 1) + 3(5x + 2) = x - 1[/tex]

[tex]\qquad❖ \: \sf \:6x + 2 + 15x + 6 = x - 1[/tex]

[tex]\qquad❖ \: \sf \:(6x + 15x - x) + (2 + 6 + 1) = 0[/tex]

[tex]\qquad❖ \: \sf \:20x + 9 = 0[/tex]

[tex]\qquad❖ \: \sf \:20x = - 9[/tex]

[tex] \qquad \large \sf {Conclusion} : [/tex]

[tex]\qquad❖ \: \sf \:x =- \cfrac{ 9}{20} [/tex]

[tex]\huge\underline{\red{A}\green{n}\blue{s}\purple{w}\pink{e}\orange{r} →}[/tex]

Step-by-step explanation:

________________________________

→ 2(3x + 1) + 3(5x + 2) = x – 1

→ 6x + 2 + 15x + 6 = x – 1

→ 21x + 8 = x – 1

→ 21x – x = – 1 – 8

→ 20x = – 9

→ x = –9/20

________________________________

The resulted integral is [tex]I=\frac{8}{3} \times \frac{5 \pi}{16}=\frac{5 \pi}{6}[/tex].

In mathematics, an integral is either a number representing the region under a function's graph for a certain interval or just an added to the initial, the derivative of which is initial function (indefinite integral).

Computation of the integrals:

Step 1: We employ the equations in cylindrical coordinates.

[tex]x=r \cos \theta, y=r \sin \theta, z=z[/tex]

Thus, in cylindrical coordinate system,

E lies above the paraboloid [tex]z=r^{2}[/tex] and below the plane [tex]z=2 r \sin \theta[/tex] .

Therefore, the top part E is  [tex]z=2 r \sin \theta[/tex] is the cross-section between paraboloid and the plane.

Now, at the cross-section use, [tex]r^{2}=2 r \sin \theta[/tex] and  [tex]z=2 r \sin \theta[/tex] .

Thus, the limits are given as ;

[tex]r^{2} \leq z \leq 2 r \sin \theta \quad 0 \leq r \leq 2 \sin \theta[/tex]

Apply the limits as compute the integration;

[tex]\begin{aligned}I=\iiint_{E} z d V &=\int_{0}^{\pi} \int_{0}^{2 \sin \theta} \int_{\tau^{2}}^{2 r \sin \theta} z r d r d z d \theta \\&=\int_{0}^{\pi} \int_{0}^{2 \sin \theta}\left[\frac{z^{2}}{2}\right]_{r^{2}}^{2 r \sin \theta} r d r d \theta \\&=\frac{1}{2} \int_{0}^{\pi} \int_{0}^{2 \sin \theta}\left[4 r^{2} \sin ^{2} \theta-r^{4}\right] r d r d \theta\end{aligned}[/tex]

                       [tex]\begin{aligned}&=\frac{1}{2} \int_{0}^{\pi} \int_{0}^{2 \sin \theta}\left[4 r^{3} \sin ^{2} \theta-r^{5}\right] d r d \theta \\&=\frac{1}{2} \int_{0}^{\pi}\left[r^{4} \sin ^{2} \theta-\frac{r^{6}}{6}\right]_{0}^{2 \sin \theta} d \theta \\&=\frac{8}{3} \int_{0}^{\pi} \sin ^{6} \theta d \theta\end{aligned}[/tex]

Step 2: Now, calculate for the [tex]I_{1}=\int_{0}^{\pi} \sin ^{6} \theta d \theta[/tex].

 [tex]\begin{aligned}\sin ^{6} \theta &=\left(\sin ^{2} \theta\right)^{2} \times \sin ^{2} \theta \\&=\left[\frac{1-\cos 2 \theta}{2}\right]^{2} \times\left[\frac{1-\cos 2 \theta}{2}\right] \\&=\frac{1}{8}\left(1-2 \cos 2 \theta+\cos ^{2} 2 \theta\right)(1-2 \cos 2 \theta) \\&=\frac{1}{8}\left(1-2 \cos 2 \theta+\frac{1+\cos 4 \theta}{2}\right)(1-2 \cos 2 \theta)\end{aligned}[/tex]

         [tex]\begin{aligned}&=\frac{1}{16}(3-4 \cos 2 \theta+\cos 4 \theta)(1-2 \cos 2 \theta) \\&=\frac{1}{32}(10-15 \cos 2 \theta+6 \cos 4 \theta-\cos 6 \theta)\end{aligned}[/tex]

Further compute the value of

   [tex]\begin{aligned}I_{1} &=\int_{0}^{\pi} \sin ^{6} \theta d \theta \\&=\frac{1}{32} \int_{0}^{\pi}(10-15 \cos 2 \theta+6 \cos 4 \theta-\cos 6 \theta) d \theta \\&=\frac{1}{32}\left[10 \theta-\frac{15 \sin 2 \theta}{2}+\frac{3 \sin 4 \theta}{2}-\frac{\sin 6 \theta}{6}\right]_{0}^{\pi} \\&=\frac{5 \pi}{16}\end{aligned}[/tex]

Therefore, the obtained integral is  [tex]I=\frac{8}{3} \times \frac{5 \pi}{16}=\frac{5 \pi}{6}[/tex].

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The complete question is -

Use cylindrical or spherical coordinates, whichever seems more appropriate. Evaluate ∫∫∫E z dV, where E lies above the paraboloid

z = x² + y²

and below the plane z = 2y. Use either the Table of Integrals or a computer algebra system to evaluate the integral.

Use the rules of exponents to simplify the expression.

[tex]\large\displaystyle\text{$\begin{gathered}\sf \frac{g^{5}h^{4} }{g^{2}h^{3} } \end{gathered}$}[/tex]

To divide powers with the same base, subtract the exponent in the denominator from the exponent in the numerator.

Subtract 2 from 5.

Subtract 3 from 4.

For any term t, t¹ =t.

Therefore, the correct option is "B".

The correct option is A. hot water moving through a pipe.

Electrical current passing through a metal wire is similar to hot water moving through a pipe.

Electrical charge carriers, often electrons or atoms deficient in electrons, travel as current. The capital letter I is a typical way to represent current. The ampere, denoted by the letter A, is the common unit.

The method of flowing of electrical current in metal wire is-

Comparison of flow of electricity with flow of water in pipe-

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The complete question is-

which of the following situations would be mist similar to an electrical current passing through a metal wire in a closed circuit?

A. hot water moving through a pipe

B. an automobile traveling through a tunnel

C. an airplane flying through clouds

D. a flying pan heating up on a hot plate

E. a bird building a nest

The mean could be used to describe a data set by itself, while interquartile range would almost never be used to describe a data set by itself.

Brief Description of Mean and Interquartile Range

While interquartile range only assesses the middle half of the data, mean and range deal with the entire set of data. The mean is significant in statistics because it helps us determine where a dataset's "center" is. The mean contains information from each observation in a dataset as a result of how it is calculated.

The interquartile range in descriptive statistics reveals the spread of your distribution's middle half. Any distribution that is sorted from low to high is divided into four equal portions using quartiles. The second and third quartiles, or the center half of your data set, are contained in the interquartile range (IQR).

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The distance from Jim's home to the gym is 12 km.

Speed is defined as the rate at which the position of object is changed.

Speed = Distance / Time

Here, there are two different speeds given for the same distance.

Average speed is the total distance travelled by the total time taken.

If distance is a constant,

Average speed = 2xy / (x + y), where x and y are two different speeds for the same distance.

Average speed = (2 × 15 × 10) / (15 + 10)

                          = 12 km/hr

So, Total distance travelled = Average speed × total time taken

                                              = 12 × 2

                                              = 24

Distance = 24 / 2 = 12

Hence, the distance from Jim's home to the gym is 12 km when average speed is taken.

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The two odd numbers are 43 and 37.

Whole numbers are positive numbers belonging to the set W ∈ {1, 2,3, 4, ...}

The two odd numbers can be represented as (2m-1) and (2n-1) respectively sum=80 and difference = 6

Let m and n be two whole numbers

Therefore

,[tex]2m-1) + (2n-1)= 80\\(2m-1) - (2n-1)= 6\\\\2(m+n-1)=80\\2(m-n)=6\\\\m+n = 41\\m-n=3\\[/tex]

Adding the two equations

[tex](m+n)+(m-n)=2m=44\\m=22\\n=m-3=19\\[/tex]

so m=22 ad n=19. our two odd numbers are 2m-1 = 43 and 2n-1 = 37

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