3 years ago

I really need help with these problems

1 answer:

4 0

10.
$x^2+14x+48=x^2+6x+8x+48=x(x+6)+8(x+6)=\boxed{(x+6)(x+8)}$

11.
$\dfrac{t^2-4t-32}{t-8}=\dfrac{t^2-8t+4t-32}{t-8}=\dfrac{t(t-8)+4(t-8)}{t-8}\\\\=\dfrac{(t-8)(t+4)}{t-8}=t+4\\\\Answer:\boxed{t+4;\ t\neq8}$

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If W(-10, 4), X(-3, -1), and Y(-5, 11) classify ΔWXY by its sides. Show all work to justify your answer.

solniwko [45]

Given:

The vertices of ΔWXY are W(-10, 4), X(-3, -1), and Y(-5, 11).

To find:

Which type of triangle is ΔWXY by its sides.

Solution:

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$WX=\sqrt{(-3-(-10))^2+(-1-4)^2}$

$WX=\sqrt{(-3+10)^2+(-5)^2}$

$WX=\sqrt{(7)^2+(-5)^2}$

$WX=\sqrt49+25}$

$WX=\sqrt{74}$

Similarly,

$XY=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(11-\left(-1\right)\right)^2}=2\sqrt{37}$

$WY=\sqrt{\left(-5-\left(-10\right)\right)^2+\left(11-4\right)^2}=\sqrt{74}$

Now,

$WX=WY$

So, triangle is an isosceles triangles.

and,

$WX^2+WY^2=(\sqrt{74})^2+(\sqrt{74})^2$

$WX^2+WY^2=74+74$

$WX^2+WY^2=148$

$WX^2+WY^2=(2\sqrt{37})^2$

$WX^2+WY^2=WY^2$

So, triangle is right angled triangle.

Therefore, the ΔWXY is an isosceles right angle triangle.

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3 years ago

Use properties to evaluate (-25) HIS O A. – 15 5 O B. 5 C. - OC. 15 O D. 1

Arisa [49]

Answer:

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Step-by-step explanation:

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2 years ago

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natta225 [31]

Answer:

Step-by-step explanation:

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3 years ago

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Find the value of y such that the distance between the given points is indicated..... (5,y) and (8,-1) is 5.

maw [93]

(5,y)(8,-1)
d = sqrt ((x2 - x1)^2 + (y2 - y1)^2)
d = sqrt ((8 - 5)^2 + (-1 - y)^2)
d = sqrt (3^2 + (-1 - 3)^2
d = sqrt (9 + (-4^2)
d = sqrt (9 + 16)
d = sqrt 25
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so ur points are : (5,-3)(8,-1)

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3 years ago

I NEED HELP, PLEASE!

aleksandrvk [35]

Answer: Irregular Hexagon

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