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2 years ago

ASAP HW HELP. what’s the answer please I don’t understand!!

Mathematics

1 answer:

irakobra [83]2 years ago

4 0

Answer:

1/2+1.5

I think. I'm sorry if I'm wrong

Step-by-step explanation:

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Find the simplified quotient. <br><br> y^2+y/y^2-2y/y^2-4y+4

BabaBlast [244]

Answer: y2-1/y-4y+4

if this helped please mark brainliest

7 0

3 years ago

I NEED HELP ASAP!!<br> like right now plz I will do anything

shepuryov [24]

Answer:

Number 5. = 69

Number 8. = 19

Step-by-step explanation:

3 0

2 years ago

A figure is composed of a semicircle and a right triangle. Determine the area of the shaded region. Use 3.14 for π and round to

Aliun [14]

Answer:

$A=9.5\ ft^2$

Step-by-step explanation:

we know that

The area of the shaded region is equal to the area of triangle plus the area of semicircle

$A=\frac{1}{2}bh+\frac{1}{2}\pi r^{2}$

we have

$b=4\ ft$

Find the height of the right triangle applying the Pythagorean Theorem

$5^2=h^2+4^2$

$h^2=9\\h=3\ ft$

The radius of the semicircle is half the height of triangle

$r=3/2=1.5\ ft$

substitute in the formula

$A=\frac{1}{2}(4)(3)+\frac{1}{2}(3.14)(1.5)^{2}$

$A=6+3.5=9.5\ ft^2$

5 0

3 years ago

Read 2 more answers

The equation tan^2 x+1=sec^2 x is an identity true or false

Solnce55 [7]

Answer:

tan²x + 1 = sec²x is identity

Step-by-step explanation:

* Lets explain how to find this identity

∵ sin²x + cos²x = 1 ⇒ identity

- Divide both sides by cos²x

∵ sin x ÷ cos x = tan x

∴ sin²x ÷ cos²x = tan²x

- Lets find the second term

∵ cos²x ÷ cos²x = 1

- Remember that the inverse of cos x is sec x

∵ sec x = 1/cos x

∴ sec²x = 1/cos²x

- Lets write the equation

∴ tan²x + 1 = 1/cos²x

∵ 1/cos²x = sec²x

∴ than²x + 1 = sec²x

- So we use the first identity sin²x + cos²x = 1 to prove that

tan²x + 1 = sec²x

∴ tan²x + 1 = sec²x is identity

8 0

3 years ago

Lets see who is smart lol

Simora [160]

9514 1404 393

Answer:

$\square\quad{y=\dfrac{1}{2}x+4}\\\\\square\quad{\dfrac{y-5}{x-2}=\dfrac{1}{2}}$

Step-by-step explanation:

The slope of a line is the same everywhere, so an equation can be written making use of that fact.

$\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}\\\\\dfrac{y-5}{x-2}=\dfrac{7-5}{6-2}=\dfrac{2}{4}\\\\\boxed{\dfrac{y-5}{x-2}=\dfrac{1}{2}}$

Cross-multiplying gives ...

2(y -5) = x -2

2y -10 = x -2 . . . eliminate parentheses

2y = x +8 . . . . . . . add 10; next, divide by 2

$\boxed{y=\dfrac{1}{2}x+4}$

These equations match choices B and D.

6 0

3 years ago