3 5 157 5 4 20 2 4 - 2 2- = 2 5 20 (Type wholebes hact fractions, or mi​

Answer:

58

Step-by-step explanation:

[tex]x = \frac{141 - 25}{2} = 58[/tex]

Answer:

Step-by-step explanation:

Since diameter x pi=3.14,this should be easy.

4x3=12

4x14/100=.56

12+.56=12.56

12.56 is the answer

Jakobe sold 150 coffee, 50 tea and 120 donuts at the coffee cart on Monday.

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the amount of coffee, y represent the amount of tea and z represent the amount of donuts, hence:

x + y + z = 320    (1)

Also:

1.5x + 2y + 0.75z = 415     (2)

And:

x = 3y     (3)

From the equations:

x = 150, y= 50 and z = 120

Jakobe sold 150 coffee, 50 tea and 120 donuts at the coffee cart on Monday.

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Answer: 51.4°

Step-by-step explanation:

Use law of cosines:

(because you are trying to find angle B, you need to use Cos B)

Cos B = c² + a²-b²/ 2ca

Then, plug in your numbers:

 Cos B = 30²+64²-51²/ 2(30)(64)

Simplify:

Cos B = 0.6237

Next, to get rid of Cos B, so that we have just B, you need to do the Arccosine or 0.6237:

B = arccosine (0.6237)

Which gets you: 51.4°    

The measure of angle B is 51.41 degrees after applying the cos law of the triangle.

In terms of geometry, the triangle is a three-sided polygon with three edges and three vertices. The triangle's interior angles add up to 180°.

We have given a triangle in the figure:

To find the measure of the angle B

We can apply cos law:

c² = a² + b² - 2ab cosB

The a, b, and c represents the side lengths of the triangle and the measures are:

a = 30

b = 64

c = 51

51² = 30² + 64² - 2(30)(64) cosB

3840CosB = 4996 - 2601

3840CosB = 2395

CosB = 2395/3840

CosB = 0.623

B = Cos⁻¹(0.623)

B = 51.41 degrees

Thus, the measure of angle B is 51.41 degrees after applying the cos law of the triangle.

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Answer:

2x+2

Step-by-step explanation:

x²+2x-3

=2(x^2-1)+2*1(2x^1-1)

Answer: 36

Step-by-step explanation:

As this is a two-digit number, it will have a digit in its 10's place and a digit in its 1's place. We can represent the 10's digit using x and the 1's digit using y.

Since the ones is twice the number of tens, this means that y is equal to 2*x. Let's put this into an equation.

[tex]y=2x[/tex]

If the 10's digit is x and the 1's digit is y, the original number would be [tex]10x + y[/tex], as x must be multiplied by 10 to get it to the 10's place. Similarly, the reversed number would be [tex]10y + x[/tex], as y must be multiplied by 10 to get it to the 10's place.

Twice the original number (10x+y) is equal to 9 more than (i.e. 9 plus) the reversed number. We can put this into an equation to help us answer the question.

[tex]2(10x+y)=9+10y+x\\20x+2y=9+10y+x\\19x+2y=9+10y\\19x=9+8y[/tex]

Now we have a system of two equations.

[tex]y=2x\\19x=9+8y[/tex]

We can solve this system by substituting y for 2x in the second equation. Then, we can isolate and get the value of x.

[tex]19x=9+8(2x)\\19x=9+16x\\3x=9\\x=3[/tex]

Now that we have the value of x, let's put it back into the first equation and solve for y.

[tex]y=2(3)\\y=6[/tex]

Remember that x is the 10's digit and y is the one's digit of our answer. Since x is 3 and y is 6, our answer is 36.

We can quickly check if our answer is right by making sure both conditions in the question are met.

6 (the one's digit) is twice the value of 3 (the ten's digit), making the first condition true. Twice of 36 is 72, which is 9 more than the reversed number, which is 63.

Answer:

this mean in case of multiplying exponents with same base you should add the exponent and keep base same.

example:

[tex] {6}^{4} \times {6}^{5 } = {6}^{4 + 5} \\ = {6}^{9} [/tex]

The ratio between 28,000 and 14 is 2000 : 1

The numbers are given as:

28,000 and 14

Express as ratio

Ratio = 28000 : 14

Divide each number by 14

Ratio = 2000 : 1

Hence, the ratio between 28,000 and 14 is 2000 : 1

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In the given diagram, the value of JI is 6

From the question, we are to determine JI

From the given diagram, we can write that

JI + IH = x + 22

Also,

JI = 2x + 28

and

IH = 5

Then, we can write that

2x + 28 + 5 = x + 22

Collect like terms

2x - x = 22 -28 -5

x = -11

Now, substitute the value of x into the equation

JI = 2x + 28

JI = 2(-11) + 28

JI = -22 +28

JI = 6

Hence, the value of JI is 6

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Answer:

[tex]A=\frac{\sqrt{3}}{4}a^2[/tex]

Step-by-step explanation:

Since an equilateral has all of the sides equal, we can find the height of triangle using: [tex]a^2+b^2=c^2[/tex]. I attached a diagram which should explain how I got the dimensions of the three sides. Using the information from the diagram we get the equation:

[tex]h^2+(\frac{a}{2})^2=a^2[/tex]

Subtract a^2 from both sides

[tex]h^2=a^2-(\frac{a}{2})^2[/tex]

Take the square root of both sides

[tex]h = \sqrt{a^2-(\frac{a}{2})^2}[/tex]

If you know the area of a triangle, it's: [tex]\frac{1}{2}bh[/tex]. In this case the base=a, and the height is what we defined above. Using this we get:

[tex]A = \frac{a}{2}*\sqrt{a^2-(\frac{a}{2})^2}[/tex]

We can distribute the exponent over the division to get:
[tex]A = \frac{a}{2}*\sqrt{a^2-(\frac{a^2}{4})[/tex]

Now we can rewrite a^2 as 4a^2/4

[tex]A = \frac{a}{2}*\sqrt{\frac{4a^2}{4}-(\frac{a^2}{4})[/tex]

Now add the two fractions:

[tex]A = \frac{a}{2}*\sqrt{\frac{3a^2}{4}[/tex]

We can distribute the square root the division just like how we distributed the exponent 2, since the square root can be expressed as an exponent (1/2)

[tex]A = \frac{a}{2}*\frac{\sqrt{3a^2}}{\sqrt{4}}[/tex]

There's a radical identity that states: [tex]\sqrt[n]{a} * \sqrt[n]{b} = \sqrt[n]{a*b}[/tex]. We can use this to rewrite one radical as multiple radicals to simplify it:

[tex]A = \frac{a}{2}*\frac{\sqrt{a^2}*\sqrt{3}}{2}[/tex]

Simplify:

[tex]A = \frac{a}{2}*\frac{a*\sqrt{3}}{2}[/tex]

Now multiply the two fractions

[tex]A = \frac{a^2*\sqrt{3}}{4}[/tex]

This is the formula for the area of an equilateral triangle, but it is also often written as:
[tex]A=\frac{\sqrt{3}}{4}a^2[/tex]

Answer:

f(n)={n/2     = if n is even=   n={2/2

       {3n+1                                 {3(2)+1   =7

Step-by-step explanation:

if n is even u can put any even no in that place and u can get odd no or even no.(i think)

the expression simplified is:

[tex]12x^3 - 25x^2 + 9[/tex]

Here we have the expression:

[tex](4x - 3)*(3x^2 - 4x - 3)[/tex]

We just need to distribute the product, we will get:

[tex](4x - 3)*(3x^2 - 4x - 3)\\\\(4x)*(3x^2 - 4x - 3) - 3*(3x^2 - 4x - 3)\\\\12x^3 - 16x^2 - 12x - 9x^2 + 12x + 9[/tex]

Now we just need to group terms with the same exponent of x:

[tex]12x^3 - 16x^2 - 12x - 9x^2 + 12x + 9\\\\12x^3 + (-16 - 9)x^2 + (-12 + 12)x + 9\\\\12x^3 - 25x^2 + 9[/tex]

That is the expression simplified

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5+a-4√(a+1)/(a-3) is the simplified form of the number. This can be obtained by multiplying numerator and denominator with conjugate of the denominator.

Given the number,

√a+1-2/√a+1+2

Denominator = √a+1+2

Conjugate of the denominator = √a+1 - 2

Now, (√a+1-2)(√a+1 - 2 )/√(a+1+2)(√a+1 - 2 )

=(√a+1 - 2 )²/(√a+1)² - 4

=(a+1 -4√a+1+5)/(a+1-4)

=5+a-4√(a+1)/(a-3)

Hence 5+a-4√(a+1)/(a-3) is the simplified form of the number.

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The equations which represents the minimum and maximum scores of the students is; x = 87 ± 1.8.

According to the task content, the

Average score of the students in the science class is; 87 points

All students are within 1.8 points of the average,

it therefore follows from the task content that the

minimum score is;

= 87 -1.8

= 85.2

maximum score is;

= 87+1.8

= 88.8 points.

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Answer:

Step-by-step explanation:

A. |x − 87| = 1.8

-hope this helps.

The domain of the rational function is:

{x ∈ ℝ| x ≠ –2, 2}

Here we have the rational function:

[tex]f(x) = \frac{x^2 - x - 12}{x^3 - 4x^2 - 4x + 16}[/tex]

We want to get the domain of that function. First, we can rewrite the numerator and denominator as:

[tex]x^2 -x - 12 = (x + 3)*(x - 4)[/tex]

[tex]x^3 - 4x^2 - 4x + 16 = (x - 4)*(x + 2)*(x - 2)[/tex]

Then we can rewrite the rational function as:

[tex]f(x) = \frac{(x + 3)*(x - 4)}{(x - 4)*(x + 2)*(x - 2)}[/tex]

This can be simplified to:

[tex]f(x) = \frac{(x + 3)}{(x + 2)*(x - 2)}[/tex]

Now, the domain will be the set of all real numbers, minus the values of x that generate problems.

In this case, the values:

x = -2 and x = 2 make the denominator to be zero, and we can't divide by zero, so  we conclue that the domain is:

{x ∈ ℝ| x ≠ –2, 2}

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Answer:

12

Step-by-step explanation:

Usually, above sea level is represented by a positive number and below sea level by a negative number.

12 ft above sea level = 12

Answer:

$4.50

Step-by-step explanation:

We can use the distributive property to solve this problem. 0.05(50+40) is the total sales tax. 0.05 being the percentage and 50 and 40 being the things we need to distribute 0.05 to. This becomes 2.5+2. This is 4.5.

Answer:

$0.45

Step-by-step explanation:

0.05(50 + 40) = 50 x 0.05 + 40 x 0.05 = 0.25 + 0.20 = 0.45

Answer:

It is an open circle and you would shade to the right on a number line.

Step-by-step explanation:

The circle is open because you are not including 3.  Put numbers in for t and you will see that the numbers greater than 3 are all on the right side of the number line.

Answer: [tex](3, \infty)[/tex]

Step-by-step explanation:

q(x) is increasing for [tex]x > 3[/tex].

p(x) is increasing for [tex]x \geq -2[/tex].

So, both functions are increasing for [tex](3, \infty)[/tex].

The cube can contain 1000 cm^3 of water

The side length is given as:

L = 10 cm

The amount of water it can contain is calculated as;

V =L^3

So, we have:

V = 10^3

Evaluate

V = 1000

Hence, the cube can contain 1000 cm^3 of water

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1. Consult the linked question. [tex]n(A\cap B\cap W) = \boxed{3}[/tex].

2. We have

[tex]n(A\cap B) = n(A\cap B \cap W) + n(A\cap B \cap W') \\\\ \implies 11 = 3 + n(A\cap B \cap W') \\\\ \implies n(A\cap B\cap W') = 8[/tex]

Similarly we can find

[tex]n(A\cap B' \cap W) = 3[/tex]

[tex]n(A'\cap B\cap W) = 6[/tex]

[tex]n(A\cap B'\cap W') = 16[/tex]

[tex]n(A'\cap B\cap W') = 8[/tex]

[tex]n(A'\cap B'\cap W) = 12[/tex]

Then the total number of students that applied to at least two of the universities is

[tex]\underbrace{n(A\cap B\cap W') + n(A\cap B'\cap W) + n(A'\cap B\cap W)}_{\text{only 2}} + \underbrace{n(A\cap B\cap W)}_{\text{all 3}} = \boxed{20}[/tex]

3. There's a small ambiguity here. Are we interested in students that applied to zero universities? If so, there are

[tex]\underbrace{n(U\setminus(A\cup B\cup W))}_{\text{none}} + \underbrace{n(A\cap B'\cap W') + n(A'\cap B\cap W') + n(A'\cap B'\cap W)}_{\text{only 1}} = \boxed{40}[/tex]

If we want students that applied to at least 1 school, we omit the first term and get a total of 36.

4. Split this set of students into those that also applied to Wachemo and those that did not.

[tex]n(A \cap B') = n(A\cap B' \cap W) + n(A\cap B'\cap W') = \boxed{19}[/tex]

The slope of the line passing through the points (-4, -9) and (5, -9) is 0

The points are (-4, -9) and (5, -9).

The slope is calculated using:

Slope = (y2 - y1)/(x2 - x1)

So, we have:

Slope = (-9 + 9)/(5 + 4)

Evaluate

Slope = 0

Hence, the slope of the line passing through the points (-4, -9) and (5, -9) is 0

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The net number of yards lost by the cowboys football team is; 8 yards.

It follows from the task content that the number of yards gained or lost by the football team during each of the three plays is; gained 7 yards, lost 8 yards and then lost 7 more yard.

Hence, by polarising yard gain as positive and yard loss as negative; we have;

Net total = 7 - 8 -7 = -8

Hence, the team lost 8 yards in the total game.

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The largest y-value from the solution set of the nonlinear system is 3.78. (Correct choice: C)

In this question we have a system of non-linear equation formed by a radical equation and a quadratic equation, which can be solved both analytically and graphically. By equalizing the two formulas we have the following equation:

√(2 · x + 4) = x² - 5 · x + 3

√(2 · x + 4) + 13 / 4  = x² - 2 · (5 / 2) · x + 25 / 4

√(2 · x + 4) + 13 / 4 = (x - 5 / 2) ²

√(2 · x + 4) = (x - 5 / 2) ² - 13 / 4

Then, we square both sides to eliminate the radical sign:

2 · x + 4 = (x - 5 / 2)⁴ - (13 / 2) · (x - 5 / 2)² + 169 / 16

2 · x + 4 = x⁴ + 4 · x³ · (- 5 / 2) + 6 · x² · (- 5 / 2)² + 4 · x · (- 5 / 2)³ + (- 5 / 2)⁴ - (13 / 2) · [x² + 2 · (- 5 / 2) · x + ( - 5 / 2)²] + 169 / 16

2 · x + 4 = x⁴ - 10 · x³ + (75 / 2) · x² - (125 / 2) · x + 625 /16 - (13 / 2) · x² + (65 / 2) · x - 325 / 8 + 169 / 16

2 · x + 4 = x⁴ - 10 · x³ + 31 · x² - 30 · x + 9

x⁴ - 10 · x³ + 31 · x² - 32 · x + 5 = 0

The real roots of this polynomial are:

x₁ ≈ 5.152, x₂ ≈ 2.862, x₃ ≈ 1.797, x₄ ≈ 0.189

Now we evaluate f(x) at each root:

y₁ ≈ 3.782, y₂ ≈ 3.118, y₃ ≈ 2.756, y₄ ≈ 2.092

The largest y-value from the solution set of the nonlinear system is 3.78. (Correct choice: C)

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Step-by-step explanation:

1) if m(∠A)∈[0;90°), then

[tex]cos(A)=\sqrt{1-sin^2A} =\frac{3}{5};[/tex]

[tex]sin2A=2sinAcosA=2*\frac{3}{5} *\frac{4}{5}=\frac{24}{25};[/tex]

[tex]cos2A=cos^2A-sin^2A=\frac{9}{25}-\frac{16}{25}=-\frac{7}{25};[/tex]

[tex]tan2A=\frac{sin2A}{cos2A}=-\frac{\frac{24}{25}}{\frac{7}{25}}=-\frac{24}{7}.[/tex]

2) if m∠A∈[90°;180°), then

cos(A)=-0.6;

sin2A=-0.96;

cos2A=-0.28;

tan2A=-24/7.

Answer:

A. [tex]h(x)=\sqrt{x-3}[/tex]

Step-by-step explanation:

The parent function of [tex]\sqrt{x}[/tex] is translated to the left when [tex]h[/tex] is positive in the transformation [tex]\sqrt{x+h}[/tex].

If [tex]h[/tex] is negative, the graph translates towards the left with the distance equal to the value of [tex]h[/tex].

Here the graph moved 3 units towards the right. This means that [tex]h[/tex] is negative and has the value of 3.

So, plugging that into the parent function for translation, the function becomes:

[tex]h(x)=\sqrt{x-3}[/tex]

The mass of the second object exists at 3 kg.

Let, F be the Force

a be the acceleration

m be the mass

The formula of force exists F = m [tex]*[/tex] a

Since force exists constant, then:

[tex]$m_1 = 2 kg, a_1 = 12 m/s^2[/tex] subset 1 designates the first object

[tex]$m_2 = ? , a_2 = 8m/s^2[/tex] subset 2 designates the second object

To estimate the mass of the second object

A constant force indicates [tex]$ m_1 * a = m_2 * a[/tex]

The mass of second object [tex]m_2[/tex], we get

[tex]m_1 * a = m_2 * a[/tex]

[tex]$m_2 = \frac{m_1 * a}{a}[/tex]

[tex]$=\frac{2*12}{8}[/tex]

= 24/8 = 3 kg

Therefore, the mass of the second object exists at 3kg.

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This snow flake-like figure can be generated by rotating an end 60° five times around the center of the hexagon. There are two forms: (i) clockwise, (ii) counterclockwise.

Geometrically speaking, snow flakes represent regular hexagons. Regular hexagons can divided into six concentric regular triangles, whose central angles have a measure of 60°. This fractal figure can be generated by rotating 60° five times around the center.

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Answer:

it matches with equation c