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Radda [10]
2 years ago
15

Find the area of a circle with a circumference of 6.28 units ​

Mathematics
1 answer:
ad-work [718]2 years ago
8 0

Answer 3.14

Step-by-step explanation:

Area= pi*r^2

Circumference=2*pi*r

Area=C^2/4pi= 6.28^2/4*pi which ='s 3.13841

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Riley solved and equation as shown in the table
Mama L [17]
She made the mistake in step 4

She should have divided by -2 not just 2 alone
6 0
3 years ago
Kyle grates 4/8 pound of cheese for enchiladas. He grates 2/8 pound of cheese for tacos. How much cheese does Kyle grate in all?
babymother [125]

He grates 6/8 pounds of cheese in all.

For this problem, you just have to add the numerators and keep the denominators.

4/8 + 2/8 = 6/8.

Remember, if the denominator is the same, just add the numerator.

7 0
3 years ago
Read 2 more answers
If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c
dolphi86 [110]

Given:

(ax+2)(bx+7)=15x^2+cx+14

And

a+b=8

Required:

To find the two possible values of c.

Explanation:

Consider

\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}

So

\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}

And also given

a+b=8---(2)

Now from (1) and (2), we get

\begin{gathered} a+\frac{15}{a}=8 \\  \\ a^2+15=8a \\  \\ a^2-8a+15=0 \end{gathered}a=3,5

Now put a in (1) we get

\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}

When a=5 and b=3, we have

\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}

Final Answer:

The option D is correct.

31 and 41

8 0
1 year ago
Find the exact value of sin(-15)
koban [17]
1)\ \ \ sin(- \alpha )=-sin \alpha \\2)\ \ \ cos2 \alpha =1-2sin^2 \alpha \ \ \ \Rightarrow\ \ \ 2sin^2 \alpha =1-cos2 \alpha \\\\ \Rightarrow\ \ \ sin \alpha = \sqrt{\frac{\big{1-cos2 \alpha}}{\big{2}}}  \\\\\\sin(-15^0)=-sin15^0=-\sqrt{\frac{\big{1-cos30^0}}{\big{2}}}  =-\sqrt{\frac{\big{1- \frac{ \sqrt{3} }{2} }}{\big{2}}} =-\sqrt{\frac{\big{2- \sqrt{3}}}{\big{4}}} =\\\\ =- \frac{1}{2}  \sqrt{2- \sqrt{3}}
7 0
3 years ago
Solve the system of equations. y=3x-1 and y=-2x+9. PLEASE SHOW WORK
kow [346]

Since both equations equal Y, set them equal to each other and solve.


3x -1 = -2x +9

Add 2x to each side:

5x - 1 = 9

Add 1 to each side:

5x = 10

Divide both sides by 5:

x = 10/5

x = 2


Now replace x with 2 and solve for y:

y = 3(2) -1 = 6-1 = 5


x = 2, y = 5  (2,5)

7 0
3 years ago
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