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notka56 [123]
3 years ago
12

Whats the area of this? Im lost...

Mathematics
1 answer:
Vlad1618 [11]3 years ago
8 0
Add the b1 with b2, which is 26. Then, multiply it by the height, which is 9. 26x9=234. Then, divide it by 2, which would result in 117. just turn it 90 degrees and u will see a trapezoid.
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Find the circumference of a circle with radius, r = 10.5m.Give your answer in terms of π<br> .
forsale [732]

Answer:

Circumference of circle =2πr=2π×10.5=21πm

Step-by-step explanation:

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3 years ago
JUST HELP ME! PLEASE!!..0o0<br><br>Applying the Pythagorean theorem, solve this triangle.​
Inessa [10]
<h3>Answer:    <u>100 ft</u>  is the hypotenuse</h3>

===================================================

Work Shown:

a = 60 and b = 80 are the two known sides

c  is the unknown hypotenuse

Applying the pythagorean theorem gets us...

a^2 + b^2 = c^2

60^2 + 80^2 = c^2

3600 + 6400 = c^2

10,000 = c^2

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c = sqrt(10,000)

c = 100

A quick way to see that 100 ft is the missing side is to note that a 6,8,10 right triangle scales up to 60,80,100 after multiplying all three sides by 10.

Or you could start with a 3,4,5 right triangle and multiply all three sides by 20.

4 0
3 years ago
Read 2 more answers
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ser-zykov [4K]
T=2 the solution set is t=2
5 0
3 years ago
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Does anyone know the answer to this one?
polet [3.4K]

Answer:

\large\boxed{a(x+4)(x-5)=0,\ a\in\mathbb{R}-\{0\}}\\\\\text{for}\ a=1\to\boxed{x^2-x-20=0}

Step-by-step explanation:

\text{If}\ x_1\ \text{and}\ x_2\ \text{are the roots of the quadratic equation}\ ax^2+bx+c=0,\\\text{then}\ ax^2+bx+c=a(x-x_1)(x-x_2).\\\\\text{If}\ x_1\ \text{and}\ x_2\ \text{, then they are the roots of the quadratic equation.}\\\\\text{We have}\ x_1=-4\ \text{and}\ x_2=5.\ \text{Therefore we have the equation:}\\\\a(x-(-4))(x-5)=0\\\\a(x+4)(x-5)=0\qquad\text{for any value of}\ a\ \text{except 0}.

7 0
3 years ago
Write an algebraic expression for the given statement Add 9z to the product of 3x and 7y
mixas84 [53]

Answer:

3x × 7y + 9z

Step-by-step explanation:

No explanation

Merry Christmas

3 0
2 years ago
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