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3 years ago

7

What is the area of the shaded region?

1 answer:

-BARSIC- [3]3 years ago

6 0

Answer:

25π - 24 cm²

Step-by-step explanation:

area of circle:

A = πr²

A = π(5²)

A = 25π

Area of triangle:

base = 6

height = 5 - 1 = 4

(1/2)(6 * 4) = 12

there are 2 triangles so unshaded area is 24 cm²

Area of shaded region = 25π - 24 cm²

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Circle O has a circumference of approximately 28.3 cm. Circle O with radius r is shown. What is the approximate length of the ra

Marrrta [24]

Answer:

4.5cm

Step-by-step explanation:

Circumference = 2 $\pi$ r

28.3=2 $\pi$ r

28.3/2 $\pi$ =r

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3 years ago

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tia_tia [17]

Answer:

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Step-by-step explanation:

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The sum of 1/6, 2/3, and 1/4 is

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<span>When adding up fractions, the idea is to bring them to a common denominator. In our case, the common denominator is 12. So we must amplify each fraction in order to bring its denominator to 12. 1/6 becomes 1*2/6*2 = 2/12. 2/3 becomes 2*4/3*4. 1/4 becomes 1*3/4*3. When we sum them up we get 2/12 + 8/12 + 3/13 or 13/12, which is (12+1)/12 or 12/12 + 1/12 or 1+ 1/12. So the answer is C.</span>

4 0

3 years ago

Х- а<br>x-b<br>If f(x) = b.x-a÷b-a + a.x-b÷a - b<br>Prove that: f (a) + f(b) = f (a + b)

GenaCL600 [577]

Given:

Consider the given function:

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}$

To prove:

$f(a)+f(b)=f(a+b)$

Solution:

We have,

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}$

Substituting $x=a$ , we get

$f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}$

$f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}$

$f(a)=0+a$

$f(a)=a$

Substituting $x=b$ , we get

$f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}$

$f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}$

$f(b)=b+0$

$f(b)=b$

Substituting $x=a+b$ , we get

$f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}$

$f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}$

$f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}$

$f(a+b)=\dfrac{b^2-a^2}{b-a}$

Using the algebraic formula, we get

$f(a+b)=\dfrac{(b-a)(b+a)}{b-a}$ $[\because b^2-a^2=(b-a)(b+a)]$

$f(a+b)=b+a$

$f(a+b)=a+b$ [Commutative property of addition]

Now,

$LHS=f(a)+f(b)$

$LHS=a+b$

$LHS=f(a+b)$

$LHS=RHS$

Hence proved.

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2 years ago

Determine whether the two expressions are equivalent.

Julli [10]

Answer:

They are equivalent because they equal the same, but the numbers are just flipped around. True

Step-by-step explanation:

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3 years ago