Pls help me !!! due today:(

Answer:

x=40

Step-by-step explanation:

first you simplify both sides of the equation, then you isolate the variable

40/4= 10

10 times 3/2 = 15

The last three digits of the final number are 109.

A number is an arithmetic value that is used to calculate and represent a quantity.

The digit zero will no longer be utilized in any counting numbers, according to the Leader of Zip, since it stands for nothing. There can only be counting numbers with 1 digit.

From 1 to 11,

11 is the tenth counting number.

So for the case of first 1000 numbers,

Number of permutations =1000/10 =100

Since 101 to 109 contains a zero.

So, 100+9=109

So, 1000+109=1109

Thus, the last number is 1109, and its last three digits are 1,0,9.

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This exercise involves the construction of a blueprint. To do so, we need to solve for angles using the Cos Rule. See the step by steps procedure below.

A blueprint is a masterplan that highlights how a structure will be look like when it is completed.

From the above, see attached the first simple quadrilateral QRST (Draft I) that is indicates the above information.

Next Step

We have to derive ∠T, ∠ S, and ∠R.
In order to achieve the above, we must split the quadrilateral into two triangles.

As noted in the image, the quadrilateral is bisected from R to T.

∠R and ∠T are also divided into R1 R2 and T1, T2 respectively.

Next step - Derive ∠T1 and ∠R1 .

To do the above,  we must deploy the Pythagoras Theorem. (PT)

PT requires:

TR² = QT² + QR²

TR² = (210)² + (515)²

TR² = 44,100 + 265,225

TR² =  309,325

TR = √309,325

TR = 556.17

Next step, Derive T1

We can derive T1 using the cos rule:

Cos T1 = QR/QT

= 515/556.17

Cos T1 = 0.926

Therefore

T1 = Cos⁻¹(0.926)

T1  = 22.18°

Next we derive R1

R1 = 180° - (90° + 22.18°)
R1 = 67.82°

In ΔTRS, by using Cosine Rule,

Cos T2 = [TS² + QR² - RS²]/2(TS) * (TR)

Cos T2 = [(232)² + (556.16)² - (542)²]/ 2 * (232) x (556.16)

Cost T2 = 69,373.945/258,058

Cost T2 ≈ 0.269

Therefore,

T2 = Cos⁻¹(0.269)

T2 ≈ 74.4°

Applying the same process, we can derive R2

Cos R2 = TR² + RS² - TS²

Cos R2 = [(556.16)² + (542)² - (232)²] / 2 (556.16) * (542)

Cost R2 = 0.911

R2 = Cos ⁻1 (0.911)

R2 = 24.34°

From the above, we can execute the following iterations:

∠S = 180° - (R2 + T2)

= 180 ° - (24.34 + 74.4)
∠C = 81.26°

Likewise:

∠R = R1 + R2
= 67.82 + 24.34
∠B = 92.16°

Lastly:

∠T = T1 + T2

= 22.18 + 74.40

∠T = 96.48°

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Full Question:

A client has just inherited a property. The deed is quite old, but she knows where the front two corners of the property are and that the distance between them is 210 feet. The shape is an irregular quadrilateral and the other three sides have a distance of 515 feet, 232 feet, and 542 feet in a clockwise direction from the front left corner. More importantly, it is known that the front left corner of the property is a right angle.

-Construct a blueprint of the property labeling the measures of all sides and angle of the property.

The number of prints that the probability of producing at least one quality print is 22.

Given that the probability of producing a high-quality color print is 0.10.

The binomial distribution summarizes the number of trials or observations when each trial has an equal chance of reaching a certain value. The binomial distribution determines the probability of observing a particular number of positive results in a defined number of tests.

Probability of high-quality color print (P) = 0.1

Probability of producing not high-quality color print (1-P) = 0.9

Assume the no. of prints to produce 'n' and these are produced independently say 'x' be the no. of producing quality prints, which follows a binomial distribution.

The probability function of binomial distribution is

[tex]P(X=x)\left(\begin{array}{l}n\\ x\end{array}\right)P^x(1-P)^{n-x},x=0,1,2,.....,n[/tex]

Given that

P(X≥1)≥0.90

1-P(X=0)≥0.90

1-(1-P)ⁿ≥0.90

1-(0.90)ⁿ≥0.90

(0.90)ⁿ≥0.10

Now, taking log of both sides, we get

nlog(0.90)≥log(0.10)

n(0.1053)≥2.3025

n≥21.86

n≈22

Hence, the probability of producing a high-quality color print is 0.10 and the number of prints to produce so that the probability of producing at least one quality print is larger than 0.90 is 22.

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Answer: -4, 0, 2

Step-by-step explanation:

The graph intersects the x-axis at [tex]x=-4, x=0, x=2[/tex].

For the given tree diagram of the experiment consisting of two trials, P(D) = 0.74

A method that can be infinitely repeated and has a clearly defined range of potential outcomes, or sample space, is referred to as an experiment or trial in probability theory. If there are multiple possible outcomes from an experiment, it is considered to be random; if there is just one, it is said to be deterministic.

Calculation of P(D) for the given experiment:

From the first branch of the tree, P(D) = 0.7 × 0.6

P(D) = 0.42

From the second branch of the tree, P(D) = 0.8 × 0.4

P(D) = 0.32

Hence, the probability of getting D in the experiment is,

P(D) = P(D) from first branch of the tree + P(D) from the second branch of the tree

P(D) = 0.42 + 0.32

P(D) = 0.74

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Question 6

1) [tex]\overline{AB} \cong \overline{BD}[/tex], [tex]\overline{CD} \perp \overline{BD}[/tex], O is the midpoint of [tex]\overline{BD}[/tex], [tex]\overline{AB} \cong \overline{CD}[/tex] (given)

2) [tex]\angle ABO, \angle ODC[/tex] are right angles (perpendicular lines form right angles)

3) [tex]\triangle ABO, \triangle CDO[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\overline{BO} \cong \overline{OD}[/tex] (a midpoint splits a segment into two congruent parts)

5) [tex]\triangle ABO \cong \triangle CDO[/tex] (LL)

Question 7

1) [tex]\angle ADC, \angle BDC[/tex] are right angles), [tex]\overline{AD} \cong \overline{BD}[/tex]

2) [tex]\overline{CD} \cong \overline{CD}[/tex] (reflexive property)

3) [tex]\triangle CDA, \triangle CDB[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\triangle ADC \cong \triangle BDC[/tex] (LL)

5) [tex]\overline{AC} \cong \overline{BC}[/tex] (CPCTC)

Question 8

1) [tex]\overline{CD} \perp \overline{AB}[/tex], point D bisects [tex]\overline{AB}[/tex] (given)

2) [tex]\angle CDA, \angle CDB[/tex] are right angles (perpendicular lines form right angles)

3) [tex]\triangle CDA, \triangle CDB[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\overline{AD} \cong \overline{DB}[/tex] (definition of a bisector)

5) [tex]\overline{CD} \cong \overline{CD}[/tex] (reflexive property)

6)  [tex]\triangle ADC \cong \triangle BDC[/tex] (LL)

7) [tex]\angle ACD \cong \angle BCD[/tex] (CPCTC)

Answer: -0.6

Step-by-step explanation:

The data of the customer responses would be classified as ordinal data.

Ordinal data is a category of categorical data that has a predetermined order or scale. Ordinal data, for instance, is said to have been gathered when a respondent rates their level of financial contentment on a scale from 1 to 10. There is no established scale for measuring the variance of each score in ordinal data.

Ordinal data is a categorical statistical data type when the distances between the categories are unknown and the variables have naturally occurring, ordered categories. On an ordinal scale, these data are available. By having a ranking, the ordinal scale differs from the nominal scale.

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Answer:

Step-by-step explanation:

hello :

the cannonball hit the ground when h=0

-5t² + 40t=0 means : -5t(t- 8)=0

t=8

The solutions are { -1,-11}

Lets solve the given equation:

x^2 + 11x + 121/4 = 125/4

Subtracting 125/4 from both sides:

=>x^2 + 11x + 121/4-125/4= 125/4 -125/4

=> x^2 + 11x - (-4/4) = 0

=> x^2 +11x -(-1) = 0

=>x^2 + 11 x + 1 = 0

This is a quadratic equation so we will use the determinant (b^2-4ac)

a = 1

b = 11

c = 1

b^2-4ac = 11^2-4*1*1 = 117

So this equation has two solutions:

x = (-b -/+ √(b^2-4ac) ) / 2a

x = (-11 -/+ √(117) ) / 2

x = (-11 -/+ 3√(13))/ 2

x = -0.91 or x = -10.9

x = -1 or x = -11 ( rounding off)

hence the solutions are { -1,-11}.

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The total bill plus interest on the credit card is $3,250

What is credit card?

Credit card gives its holder the opportunity to make purchases on credit such that amount spent and the interest that accrues thereon would be paid by the credit card holder at a future time.

In this case, the function shows the relationship between the amount owed and the interest so as to give the outstanding balance.

f(p) = p(1.30)

p=amount owed=$2,500

f(p)=$2,500*(1.30)

f(p)=$3,250

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Graph is in the attached image.

The rate at which Sean drank the slushy is 5 milliliters  per second.

The time it would take to drink all the slushy is 55 seconds.

In order to determine the speed at which the slushy was drank, divide the slushy drank in 13 seconds by 13 seconds.

Speed = slushy drank in 13 seconds / time

(275 - 210) / 13

65 / 13 = 5 milliliters  per second

Time it would take to drink all the slushy =total milliliters of the slushy / speed

275 / 5 = 55 seconds.

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Answer:

I don't know.i don't know

The surface area of the composite figure is 444m².

Given a composite figure which is shown in the question.

The  area of ​​a shape (in square units) is the number of unit squares required to cover the entire area without gaps or overlaps. If a hologram has planes, those faces are called faces.

Firstly, we will find the area of front and end triangles by using the formula

Area=(1/2)×b×h and we will multiply this area by 2 because we are finding the area of two same triangles.

Here, h=6 and b=16 and substitute these values in the formula, we get

A₁=2×(1/2)×16×6

A₁=2×8×6

A₁=96m²

Now, we will find the area of the left and right side rectangles which joined both the triangles.

We will find the area by using the formula Area=l×b and we will multiply this area by 2 because we are finding the area of two same rectangles.

here, l=10 and b=5 and substitute these values in the formula, we get

A₂=2×10×5

A₂=100m²

Further, we will find the area of the front and end side rectangles that joined both by the base of the triangles.

We will find the area by using the formula Area=l×b and we will multiply this area by 2 because we are finding the area of two same rectangles.

here, l=16 and b=4 and substitute these values in the formula, we get

A₃=2×16×4

A₃=128m²

Furthermore,  we will find the area of the left and right side rectangles which joined by the front and end rectangles.

We will find the area by using the formula Area=l×b and we will multiply this area by 2 because we are finding the area of two same rectangles.

here, l=4 and b=5 and substitute these values in the formula, we get

A₄=2×4×5

A₄=40m²

Now, we will find the area of the base of the composite figure which is rectangle.

We will find the area by using the formula Area=l×b.

here, l=16 and b=5 and substitute these values in the formula, we get

A₅=16×5

A₅=80m²

So, the surface area of the given composite figure will be

Surface area=A₁+A₂+A₃+A₄+A₅

Surface area=96m²+100m²+128m²+40m²+80m²

Surface area=444m²

Hence, the surface area of the given composite figure is 444m².

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Step-by-step explanation:

[tex]12 {x}^{2} - 6( {a}^{2} + {b}^{2} )x + 3 {a}^{2} {b}^{2} = 0[/tex]

Using quadratic formula, we get

[tex]x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{( - 6) {}^{2}( {a}^{2} + {b}^{2}) {}^{2} - 4(12)(3 {a}^{2} {b}^{2} )} }{24} [/tex]

[tex]x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{36( {a}^{2} + {b}^{2} ) {}^{2} - 144 {a}^{2} {b}^{2} } }{24} [/tex]

[tex]x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{36( ({a}^{2} + {b}^{2}) {}^{2} - 4 {a}^{2} {b}^{2} }) }{24} [/tex]

[tex]x = 6( {a}^{2} + {b}^{2} )± \frac{6 \sqrt{ ({a}^{2} + {b}^{2} ) {}^{2} - 4 {a}^{2} {b}^{2} } }{24} [/tex]

[tex]x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{( {a}^{2} + {b}^{2} ) {}^{2} - 4 {a}^{2} {b}^{2} } }{4} [/tex]

Answer: x₁=6b²,  x₂=6a².

Step-by-step explanation:

[tex]12x^2-6*(a^2+b^2)+3a^2b^2=0\\D=(6*(a^2+b^2))^2-4*12*3a^2b^2\\D=6^2*(a^2+b^2)^2-144a^2b^2\\D=36*(a^4+2a^2b^2+b^4)-144a^2b^2\\D=36a^4+72a^2b^2+b^4-144a^2b^2\\D=36a^4-72a^2b^2+36b^4\\D=(6a^2)^2-2*6a^2*6b^2+(6b^2)^2\\D=(6a^2-6b^2)^2\\\sqrt{D}=\sqrt{(6a^2-6b^2)^2}=|6a^2-6b^2|=б(6a^2-6b^2).\\\displaystyle x_{1,2}=\frac{-(-6*(a^2+b^2))б(6a^2-6b^2)}{2} .\\x_1=\frac{6a^2+6b^2-6a^2+6b^2}{2} \\x_1=\frac{12b^2}{2} \\x_1=6b^2.\\x_2=\frac{6a^2+6b^2+6a^2-6b^2}{2} \\x_2=\frac{12a^2}{2} \\x_2=6a^2.[/tex]

Answer:

x = - 1

Step-by-step explanation:

- 15 = 22x + 7 ( subtract 7 from both sides )

- 22 = 22x ( divide both sides by 22 )

- 1 = x

Answer:

b. x^2+4x

Step-by-step explanation:

So since you don't know the width, you can just represent it as a variable, and in this case that variable will be "x". Now using this variable we can use it to represent the length since there is a relationship between the width and length which is given. Since the length is 4 feet greater than the width, we can represent the length as (x+4). Now to find the area, we simply multiply the length * width, which is x * (x+4) = x^2 + 4x

Answer:

1) (x-6)^2=49

2) C

Step-by-step explanation:

Subtract 12 x and you get

-x^2-12x+13=0

You will be left with, after making a perfect square

(x-6)^2=49

To get rid of the squared/2nd power, add plus or minus to the opposite side and square root both sides. You will be left with

x-6=plus or minus 7

Your answer comes out to the to problem. Setting both to 0, you get -13, and 1. -6 plus 7 is 1, and -6 plus a -7 is -13.

The length of the triangle will be equal to 2.7.

It should be noted that that sin will be the opposite divided by the hypothenuse.

The measure of angle L is 20°.

This will be:

sin 20 = LN/8

NM = 8 sin 20

NM = 2.7

In conclusion, NM is 2.7.

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Answer:   2.7

Step-by-step explanation:  TRUST ME!!!  Answer is correct on Edge!!!  :)

The slope of the parallel line is 0 and the slope of the perpendicular line is undefined

The equation is given as:

y = 1

A linear equation is represented as:

y = mx + c

Where:

m represents the slope

By comparison, we have:

m = 0

The slope of the parallel line is:

Slope = m

This gives

Slope = 0

The slope of the perpendicular line is:

Slope =-1/m

This gives

Slope = -1/0

Slope = undefined

Hence, the slope of the parallel line is 0 and the slope of the perpendicular line is undefined

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When the balloon, rising at 10 ft/sec, rises to 100 feet, the rate of change of the angle of elevation is 0.05 rad/sec

The rate at which the balloon rises = 10 ft/sec

Horizontal distance of the balloon from the observer, d = 100 feet

Therefore;

[tex] \frac{dy}{dt} = 10[/tex]

The angle of elevation can be presented as follows;

[tex] tan(\theta) = \frac{y}{100} [/tex]

[tex] y = 100 \times tan(\theta) [/tex]

Therefore;

[tex] \frac{dy}{dt} = \mathbf{ 100\cdot sec^2 \frac{d \theta}{dt}} [/tex]

Which gives;

[tex] \frac{d \theta}{dt}= \frac{ \frac{d y}{dt}}{100} \times cos^2 \theta [/tex]

[tex] \mathbf{ \frac{d \theta}{dt}} = \frac{ 10}{100} \times cos^2 \theta [/tex]

When the balloon is 100 ft. from the ground, we have;

Therefore;

[tex] \frac{d \theta}{dt}= \frac{ 10}{100} \times \left(\frac{1}{\sqrt{2}} \right)^2 [/tex]

[tex] \frac{ 10}{100} \times \left(\frac{1}{\sqrt{2}} \right)^2 = \frac{ 1}{20} [/tex]

Which gives;

[tex] \frac{d \theta}{dt}=\frac{ 1}{20} = 0.05 [/tex]

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Option D. What we can conclude is that  If the end result from the second column is not 3, then the sum of the numbers in the first column is not equal to the sum of the numbers in the second column.

An operation is done on a first column of numbers. When this operation is repeated on the second column, we would have option D.

If the sum of the numbers that are in the third can be seen to be greater than three, then the sums are not going to be seen as equal

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One column of numbers consists of 28, 42, and 14. When the digits of the numbers are added together, the result is 2 + 8 + 4 + 2 + 1 + 4 = 21, and when the digits of 21 are then added together, the end result is 2 + 1 = 3. If the same process is performed on the numbers in a second column, what can be concluded?

A. If the end result from the second column is also 3, then the sum of the numbers in the first column is not equal to the sum of the numbers in the second column.

B. If the end result from the second column is not 3, then the sum of the numbers in the first column is equal to the sum of the numbers in the second column.

C. If the end result from the second column is also 3, then the sum of the numbers in the first column is equal to the sum of the numbers in the second column.

D. If the end result from the second column is not 3, then the sum of the numbers in the first column is not equal to the sum of the numbers in the second column.

The probability expression is [tex]\frac{^{6}C_2}{^{20}C_2}[/tex]

The given parameters are

The total number of people is

n = 20

2 of the 20 people are to be selected.

So, the number of selection is

[tex]n = ^nC_r[/tex]

This gives

[tex]n = ^{20}C_2[/tex]

The expression that represents both students being sophomore is

[tex]n = ^{6}C_2[/tex] --- i.e. 2 students selected from 6

The probability is then represented as:

[tex]\frac{^{6}C_2}{^{20}C_2}[/tex]

Hence, the probability expression is [tex]\frac{^{6}C_2}{^{20}C_2}[/tex]

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Answer:

a

Step-by-step explanation:

edge 2023

Answer:

Step-by-step explanation:

hello :

the perimeter of triangle is : P= (x-z+yx)+( z+x+xz)+A

given  P=x+y+xy  calculate the third side  A

x+y+xy = (x-z+yx)+( z+x+xz) so :  A=x+y+xy -x+z-yx-z-x-xz....continu

a=

If the perimeter of the triangle park is x + y + xy meter. Then the third length of the triangle will be y - x - xz meters.

The polygonal shape of a triangle has a number of sides and three independent variables. Angles in the triangle add up to 180°.

The perimeter of the triangle is determined by adding together all of its sides.

The perimeter of the triangle park is x + y + xy meter. The length of the two sides are x - z + yx meter and z + x + xz meter.

Let the third side be 'A'. Then the third length of the triangle is given as,

x + y + xy = (x - z + xy) + (z + x + xz) + A

x + y + xy = x - z + xy + z + x + xz + A

y = x + xz + A

A = y - x - xz

If the perimeter of the triangle park is x + y + xy meter. Then the third length of the triangle will be y - x - xz meters.

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Hugh bough 4 books.

What is linear equation ?

The cost of magazines =

3.95(3) + 8.95b = 47.65

11.85 + 8.95b = 47.65

Subtract 11.85 from both sides.

8.95b = 35.8

Divide 8.95 from both sides.

b = 4

Therefore, Hugh bough 4 books.

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The complete question is -

Hugh bought some magazines that cost $3.95 each and some books that cost $8.95 each. He spent a total of $47.65. If Hugh bought 3 magazines, how many books did he buy? The equation that models the problem is 3.95m + 8.95b = 47.65, where m is the number of magazines and b is the number of books.

The diagram where angles 1 and 2 form a linear pair is option D which is that a horizontal line has 1 line extending from it. Angles 1 and 2 are formed.

Given that angle 1 and 2 form a linear pair.

We know that a linear pair is an angle that are formed when two lines intersect each other at a single point.

In our case when a horizontal line has 1 line extending from it and when angles 1 and 2 are formed shows a linear pair.

In first part when two lines intersect they donot form 4 angles.

In second part when 3 lines extend from a point they don't form right angle between the lines.

Hence the linear pair angles are shown by the statement that a horizontal line has 1 line extending from it. Angles 1 and 2 are formed.

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