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4 years ago

What is the center of the circle shown below

Mathematics

2 answers:

ivanzaharov [21]4 years ago

7 0

That's the only point within the circle, and it seems like the center. In these questions, just assuming it's perfectly in the center typically works.

Vinvika [58]4 years ago

6 0

the correct answer to your question is : J

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Use the scale drawing and the scale factor to enlarge a square that has a side length of 12 in. Scale factor= 3 in:2m.

podryga [215]

Given:

The side of square = 12 in.

Scale factor of enlargement = 3 in : 2 m

To find:

The proportion that is use to solve the side length, x, of the enlarged square.

Solution:

Let, the side of length of enlarged square = x m

In case of enlargement the corresponding sides are proportional.

$\dfrac{3}{12}=\dfrac{2}{x}$

$3x=2\times 12$

$3x=24$

Divide both sides by 3.

$x=\dfrac{24}{3}$

$x=8$

Therefore, the required proportion is $\dfrac{3}{12}=\dfrac{2}{x}$ and the side length of the square after enlargement is 8 m.

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3 years ago

What is the value of k? -k+78=98-20

Jlenok [28]

Answer:

k=0

Step-by-step explanation:

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3 years ago

What is the slope of the line that passed through (3,-4) and (4,9)

Advocard [28]

Step-by-step explanation:

Step 1: Find the slope

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{9-(-4)}{4-3}$

$m =\frac{9+4}{1}$

$m=13$

Answer: The slope is 13

5 0

4 years ago

Find the coefficient of the x^3 in the expansion of (2x-9)^5

il63 [147K]

Use the binomial theorem:

$\displaystyle (2x-9)^5 = \sum_{k=0}^5 \binom5k (2x)^{5-k}(-9)^k = \sum_{k=0}^5 \frac{5!}{k!(5-k)!} 2^5 \left(-\frac92\right)^k x^{5-k}$

The x ³ terms occurs for 5 - k = 3, or k = 2, and its coefficient would be

$\dfrac{5!}{2!(5-2)!} 2^5 \left(-\dfrac92\right)^2 = \boxed{6480}$

8 0

3 years ago

Please anybody help me with this question

inn [45]

Answer:

y = 12

Step-by-step explanation:

Divide both by 1/4

x = 1 when y = 32/3

When x = 1 1/8 = 9/8

y = 32/3 x 9/8

y = 12

6 0

3 years ago