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2 years ago

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A man is 6 feet 2 inches tall. To find the height of a tree, the shadow of the man and the shadow of the tree were measured. The

length of the man's shadow was 2 feet 1 inch. The length of the tree's shadow was 3 feet 7 inches. What is the height of the tree?
A. 10.75ft
B. 10.9ft
C. 14.3ft
D. 12.9ft

1 answer:

Readme [11.4K]2 years ago

3 0

Here's my work. Hope it helps.

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What are the zeros of the quadratic function f(x) = 6x2 + 12x – 7?

zalisa [80]

we have

$f(x) = 6x^{2} + 12x -7$

To find the zeros equate the function to zero

$6x^{2} + 12x -7=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$7= 6x^{2} + 12x$

Factor the leading coefficient

$7= 6(x^{2} + 2x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$7+6= 6(x^{2} + 2x+1)$

$13= 6(x^{2} + 2x+1)$

Rewrite as perfect squares

$13= 6(x+1)^{2}$

$(13/6)=(x+1)^{2}$

square root both sides

$x+1=(+/-)\sqrt{\frac{13}{6}}$

$x=-1(+/-)\sqrt{\frac{13}{6}}$

$x1=-1+\sqrt{\frac{13}{6}}$

$x2=-1-\sqrt{\frac{13}{6}}$

therefore

the answer is

The zeros of the quadratic function are

$x=-1+\sqrt{\frac{13}{6}}$ and $x=-1-\sqrt{\frac{13}{6}}$

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2 years ago

Read 2 more answers

Eliza’s backpack weighs 18 and StartFraction 7 over 9 EndFraction pounds with her math book in it. Without her math book, her ba

xenn [34]

Answer:

3 65/72

Step-by-step explanation:

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1 year ago

In a data set with a range of 55.4 to 105.4 and 400 observations, there are 176 observations with values less than 86. Find the

IrinaVladis [17]

<u>ANSWER: </u>

In a data set with a range of 55.4 to 105.4 and 400 observations.86 lies in the 49th percentile.

<u>SOLUTION: </u>

Given, in a data set with a range of 55.4 to 105.4 and 400 observations.

There are 176 observations below the value of 86, and we need to find the percentile for 86.

We know that, percentile formula = $\frac{\text {number of observations below the required number}}{\text {total number of observations}} \times 100$

Percentile of 86 = $\frac{176}{400} \times 100$

Since, we cancelled 400 with 100 we get 4 , hence above expression becomes,

$=\frac{176}{4}$ = 49

So, percentile of 86 = 49

Hence, 86 lies in the 49th percentile.

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2 years ago

For each system of equations, drag the true statement about its solution set to the box under the system?

natta225 [31]

Answer:

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

Step-by-step explanation:

* Lets explain how to solve the problem

- The system of equation has zero number of solution if the coefficients

of x and y are the same and the numerical terms are different

- The system of equation has infinity many solutions if the

coefficients of x and y are the same and the numerical terms

are the same

- The system of equation has one solution if at least one of the

coefficient of x and y are different

* Lets solve the problem

∵ y = 4x + 2 ⇒ (1)

∵ y = 2(2x - 1) ⇒ (2)

- Lets simplify equation (2) by multiplying the bracket by 2

∴ y = 4x - 2

- The two equations have same coefficient of y and x and different

numerical terms

∴ They have zero equation

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

∵ y = 3x - 4 ⇒ (1)

∵ y = 2x + 2 ⇒ (2)

- The coefficients of x and y are different, then there is one solution

- Equate equations (1) and (2)

∴ 3x - 4 = 2x + 2

- Subtract 2x from both sides

∴ x - 4 = 2

- Add 4 to both sides

∴ x = 6

- Substitute the value of x in equation (1) or (2) to find y

∴ y = 2(6) + 2

∴ y = 12 + 2 = 14

∴ y = 14

∴ The solution is (6 , 14)

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

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2 years ago

Simplify square root of (1-sin theta)(1+sin theta)

Vinil7 [7]

Recall the important identities:

i) $(a-b)(a+b)= a^{2}- b^{2}$ , the difference of squares formula.

ii) $sin^{2} x+cos^{2}x=1$ , for any angle x.

Then,

from i)

$\sqrt{(1-sin(theta))(1+sin(theta))}= \sqrt{1- sin^{2}(theta) }$

then, from ii), we have

$\sqrt{1- sin^{2}(theta) }= \sqrt{cos^{2}(theta) }=cos(theta)$

Answer: cos(theta)

6 0

2 years ago