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Katyanochek1 [597]
2 years ago
14

If the area of a circle measures 25π what is the circumference of the circle in terms of π​

Mathematics
1 answer:
oksian1 [2.3K]2 years ago
3 0

Answer:

Explain better please i will awnser

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If f(x) = 4x - 7, what's the value of f(2) ? *<br> O -1<br> O 1<br> O 20<br> O 35
sp2606 [1]

Answer:

B

Step-by-step explanation:

f(x) = 4x - 7

What does f(x) mean? It means that whatever variable you have on the left which is (x), the same variable must be on the right.

So f(2) means that whatever variable is on the right, it must be replaced with 2 on the right. So .....

f(2) = 4(2) - 7

f(2) = 8 - 7

f(2) = 1

3 0
2 years ago
Quadrilateral ABCD is located at A (−2, 2), B (−2, 4), C (2, 4), and D (2, 2). The quadrilateral is then transformed using the r
Mademuasel [1]
New cordinates are formed by adding 7 in x and subtracting 2 from y
A(−2, 2) =A ' (-2 +7 , 2 - 1 ) = A' (5,1)
B(−2, 4)  = B' (-2 + 7 , 4 -1 )= B' (5,3)
C(2, 4)   =  C' (2 + 7 , 4 -1 )= C' (9,3)
<span>D(2, 2)   D' (2 + 7 , 2 -1 ) = D' ( 9 , 1)</span><span>

</span>
8 0
2 years ago
A rectangle has a width one less than the length. If the perimeter is 30 units find the width and length
Olegator [25]

Answer:

7 units by 8 units

Step-by-step explanation:

ALL POSSIBLE PERIMETERS:

1x14

2x13

3x12

4x11

5x10

6x9

7x8

5 0
3 years ago
What is the solution to x + 12 = 25?<br>A. x=13<br>B. x=12<br>C. x=37<br>D. x=24​
Archy [21]

Answer:

A . x=13 is the solution of this question.

6 0
2 years ago
Read 2 more answers
Consider an urn containing 8 white balls, 7 red balls and 5 black balls.
weqwewe [10]

Answer + Step-by-step explanation:

1) The probability of getting 2 white balls is equal to:

=\frac{8}{20} \times \frac{7}{19}\\\\= 0.147368421053

2) the probability of getting 2 white balls is equal to:

=C^{2}_{5}\times (\frac{8}{20} \times \frac{7}{19}) \times (\frac{12}{18} \times \frac{11}{17} \times \frac{10}{16})\\=0.397316821465

3) The probability of getting at least 72 white balls is:

=C^{72}_{150}\times \left( \frac{8}{20} \right)^{72}  \times \left( \frac{7}{20} \right)^{78}  +C^{73}_{150}\times \left( \frac{8}{20} \right)^{73}  \times \left( \frac{7}{20} \right)^{77}  + \cdots +C^{149}_{150}\times \left( \frac{8}{20} \right)^{149}  \times \left( \frac{7}{20} \right)^{1}  +\left( \frac{8}{20} \right)^{150}

=\sum^{150}_{k=72} [C^{k}_{150}\times  \left( \frac{8}{15} \right)^{k}  \times \left( \frac{7}{15} \right)^{150-k}]

5 0
11 months ago
Read 2 more answers
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