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3 years ago

SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!

1 answer:

5 0

(2y² + 7y + 11) - (8y² - 5y + 7) you can distribute the negative sign to the second expression and then combine the like terms

new expression: (2y² + 7y + 11) (-8y² + 5y - 7)

2y² and -8y² equals -6y²

7y and 5y equals 12y

11 and -7 equals 4

The answer is -6y² + 12y + 4

You might be interested in

If tan theta = 10/13 and cos theta > 0 then sin2theta is what ?

svetoff [14.1K]

$sin2\alpha = \frac{260}{269}$

<u>Step-by-step explanation:</u>

We have , $Tan\alpha = \frac{Perpendicular}{Base} = \frac{10}{13}$ ,

We know that $sin\alpha = \frac{Perpendicular}{Hypotenuse} = \frac{Perpendicular}{\sqrt[2]{(Perpendicualr)^{2} + (Base)^{2})} }$

Substituting values of P & B , $sin\alpha = \frac{10}{\sqrt{10^{2} + 13^{2}} } = \frac{10}{\sqrt{269} }$

Now , $sin2\alpha = 2sin\alpha cos\alpha = 2sin\alpha \sqrt{1 - (sin\alpha)^{2} }$

⇒ $sin2\alpha =$ $\frac{10}{\sqrt{269} }$ × $\sqrt{1 - (\frac{10}{\sqrt{269} })^{2} }$ ×2

⇒ $sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{269 - 100}{269} } )$

⇒ $sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{169 }{269} } )$

⇒ $sin2\alpha = \frac{260}{\sqrt{269} }( \sqrt{\frac{1}{269} } )$

⇒ $sin2\alpha = \frac{260}{269}$

6 0

3 years ago

Alex rode his bike 25 kilometers. About how many miles did Alex ride his bike?

Advocard [28]

Answer:

Alex rode 15.5343 miles in his bike

Step-by-step explanation:

1 kilometer is 0.621371 miles, multiply 0.621371 with 25. Hope this helps!

6 0

3 years ago

The vertices of ΔGHI are G(-3,0), H(-5,-5), and I(1,-5). What are the coordinates of the image after being reflected over the y-

s2008m [1.1K]

After reflecting the points of triangle GHI over the y axis, the new points will be:
G' (3,0)
H' (5,-5)
I' (-1,-5)

6 0

3 years ago

the altitude of the hypotenuse of a right triangle divides the hypotenuse into segments of lengths 14 and 8. what is the length

lisov135 [29]

Answer:

B. $4\sqrt{7}$

Step-by-step explanation:

The right triangle altitude theorem states that the altitude of a right angled triangles formed on the hypotenuse is equal to the geometric mean of the 2 line segments it creates.

This can be represented as:

$h = \sqrt{(xy)}$

Where,

h = the length of the altitude,

x and y are the lengths of the 2 segments formed.

Therefore, the length of the altitude = $h = \sqrt{(14*8)}$

$h = \sqrt{112}$

$h = \sqrt{16}*\sqrt{7}$

$h = 4\sqrt{7}$

3 0

4 years ago

When an object with a momentum of 80 kg ´ m/s collides with an object with a momentum of –100 kg ´ m/s the total momentum after

mojhsa [17]

This is a physics problem.

The law is that the momentum is preserved: total momentum before collsion = total momentum after collision

Momentum before collision = 80 kg.m/s - 100 kg.m/s = -20 kg.m/s

Then answer is b. - 20 kg.m/s

3 0

3 years ago