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3 years ago

9

The scale for the drawing of a rectangular playing field is 2 inches=7 feet. Find an equation you can use to find the dimensions

of the actual field. What are the actual dimensions?

1 answer:

lina2011 [118]3 years ago

5 0

Answer:

The actual answer is 7/2

Step-by-step explanation:

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What is the value of the number 6 in the total number 4560 thousand

julsineya [31]

The value of 6 is 60, since it is in the tens place. Hope this helps!

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$\bf |x^2+3x-1|\ \textless \ 3\implies 
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3 years ago

there are 6 chocolate cupcakes and 4 strawbwrry cuocakea. write a number sentence to compare chocolate cupcakes to strawberry cu

Pepsi [2]

Try to remember that thee alligator wants to eat more chocolate cupcakes

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3 years ago

Y=6x+20 substitution

Oksana_A [137]

Answer:

-10/3 or -3.33

Step-by-step explanation:

substitute 0 for y

0 = 6x + 20

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-20/6 = 6x/6

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3 years ago

In Triangle XYZ, measure of angle X = 49° , XY = 18°, and

marissa [1.9K]

Answer:

There are two choices for angle Y: $Y \approx 54.987^{\circ}$ for $XZ \approx 15.193$ , $Y \approx 27.008^{\circ}$ for $XZ \approx 8.424$ .

Step-by-step explanation:

There are mistakes in the statement, correct form is now described:

<em>In triangle XYZ, measure of angle X = 49°, XY = 18 and YZ = 14. Find the measure of angle Y:</em>

The line segment XY is opposite to angle Z and the line segment YZ is opposite to angle X. We can determine the length of the line segment XZ by the Law of Cosine:

$YZ^{2} = XZ^{2} + XY^{2} -2\cdot XY\cdot XZ \cdot \cos X$ (1)

If we know that $X = 49^{\circ}$ , $XY = 18$ and $YZ = 14$ , then we have the following second order polynomial:

$14^{2} = XZ^{2} + 18^{2} - 2\cdot (18)\cdot XZ\cdot \cos 49^{\circ}$

$XZ^{2}-23.618\cdot XZ +128 = 0$ (2)

By the Quadratic Formula we have the following result:

$XZ \approx 15.193\,\lor\,XZ \approx 8.424$

There are two possible triangles, we can determine the value of angle Y for each by the Law of Cosine again:

$XZ^{2} = XY^{2} + YZ^{2} - 2\cdot XY \cdot YZ \cdot \cos Y$

$\cos Y = \frac{XY^{2}+YZ^{2}-XZ^{2}}{2\cdot XY\cdot YZ}$

$Y = \cos ^{-1}\left(\frac{XY^{2}+YZ^{2}-XZ^{2}}{2\cdot XY\cdot YZ} \right)$

1) $XZ \approx 15.193$

$Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-15.193^{2}}{2\cdot (18)\cdot (14)} \right]$

$Y \approx 54.987^{\circ}$

2) $XZ \approx 8.424$

$Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-8.424^{2}}{2\cdot (18)\cdot (14)} \right]$

$Y \approx 27.008^{\circ}$

There are two choices for angle Y: $Y \approx 54.987^{\circ}$ for $XZ \approx 15.193$ , $Y \approx 27.008^{\circ}$ for $XZ \approx 8.424$ .

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3 years ago