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Oduvanchick [21]
2 years ago
7

What is the circumference of the circle? 11 m​

Mathematics
1 answer:
-BARSIC- [3]2 years ago
3 0
C=2πr r radius it would be right
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Graph the solution set.-10x≤2y
vagabundo [1.1K]

To be able to determine the graph of this inequality, we'll start rearranging the inequality putting the "y" variable at the left side of the equation.

\begin{gathered} -10x\leq2y \\ \frac{-10x}{2}\leq\frac{2y}{2} \\ -5x\leq y \\ or \\ y\ge-5x \end{gathered}

Since the inequality here is greater than or equal to, this means that the shade is above the solid line.

This equation also has a slope of -5 and y-intercept of 0.

Therefore, the graph of this equation looks like this:

4 0
1 year ago
Anybody good in geometry!?
crimeas [40]

7th:

Y + 128° = 180° (Linear Pair)

Y = 180 - 128 = <em>52</em><em>°</em>

<em>You're</em><em> </em><em>welcome</em><em>.</em><em>.</em><em>.</em><em>Please</em><em> </em><em>mark</em><em> </em><em>as</em><em> </em><em>brainliest</em><em> </em>

<em>au revoir madame</em>

5 0
2 years ago
Read 2 more answers
Three roads, A, B and C, all run south to north. They each meet a highway that runs west to east. Road B is 4.6 mi east of Road
WITCHER [35]

Answer:

Road C is 1.5 mi east of Road A

Step-by-step explanation:

we know that

Road B is 4.6 mi east of Road A

Road C is 3.1 mi west of Road B

so

Road C is between Road A and Road B

AB=4.6 miles

CB=3.1 miles

AB=AC+CB

Solve for AC

AC=AB-CB

AC=4.6-3.1=1.5 miles

therefore

Road C is 1.5 mi east of Road A

7 0
3 years ago
Please help me. This is real confusing. ​
katrin2010 [14]

Answer:

1st problem: b) A=2500(1.01)^{12t}

2nd problem:  c) A=2500e^{.12t}

Step-by-step explanation:

1st problem:

The formula/equation you want to use is:

A=P(1+\frac{r}{n})^{nt}

where

t=number of years

A=amount he will owe in t years

P=principal (initial amount)

r=rate

n=number of times the interest is compounded per year t.

We are given:

P=2500

r=12%=.12

n=12 (since there are 12 months in a year and the interest is being compounded per month)

A=2500(1+\frac{.12}{12})^{12t}

Time to clean up the inside of the ( ).

A=2500(1+.01)^{12t}

A=2500(1.01)^{12t}

----------------------------------------------------

2nd Problem:

Compounded continuously problems use base as e.

A=Pe^{rt}

P is still the principal

r is still the rate

t is still the number of years

A is still the amount.

You are given:

P=2500

r=12%=.12

Let's plug that information in:

A=2500e^{.12t}.

6 0
3 years ago
29 less than -10 times a number is equal to -18 times a number +91
sergejj [24]

Answer:

10

Step-by-step explanation:

5 0
3 years ago
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