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3 years ago

Of the balls in Brian's garage, 1/3 are basketballs and 1/15 are soccer balls. 10 points

Mathematics

1 answer:

makkiz [27]3 years ago

8 0

Answer:

8/15 balls

Step-by-step explanation:

You add the fraction of basketballs to the fraction of soccer balls. 1/3+1/5=8/15. Hope this helps.

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. The concert tickets were $57 plus two souvenirs for your friend and yourself. You spent $94

Lena [83]

Maybe $47 for each souvenir

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3 years ago

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Sam Is 11 years older than Susan the sum of their ages in years is 27 what’s sams age in years

sweet [91]

Answer:

Step-by-step explanation:

Let Sam be x

Let Susan be y

x = y+11

x + y = 27

Substitute (x) with (y+11)

(y+11) + y = 27

Solve the equation

2y + 11 = 27

2y = 16

y = 8

Susan's age is 8.

Sam's age is 8 + 11

Sam's age is 19

Hope this helped :)

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3 years ago

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What is the best approximation for the circumference of a circle with a diameter of 300 feet

garri49 [273]

Answer:

The answer is C=942.48

Step-by-step explanation:

You would figure this out by using the formula C=2πr.

So let's break this down.

Circumference (C) equals (=) pi (π) times 2 times the radius of the circle.

Since the problem gave you the diameter which is the length from one side of the circle to the other, you have to divide the diameter in half. Which would give you a radius of 150.

Taking the 150 you would multiply 10*2*π.

Which would give you....(drumrollllll)

C=942.48

I hope this helps!!

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3 years ago

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I don't understand the problem

Rasek [7]

The answer is D.
18 2/3÷2/3
18.66÷2/3
27.99
28

5 0

3 years ago

Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)

Leni [432]

Alrighty

remember some rules
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}=\frac{1}{x^m}$
and
$x^0=1$ for all real values of x
and
$(a^b)^c=a^{bc}$
and
$(\frac{a}{b})^c=\frac{a^c}{b^c}$
and
(a/b)/(c/d)=(ad)/(bc)
and
$(a^b)(a^c)=a^{b+c}$
and
$\frac{a^m}{a^n}=a^{m-n}$
and don't forget pemdas
example: $-x^m=-1(x^m)$ but $(-x)^m=(-1)^m(x^m)$

so

4.
$(\frac{c^{-2}}{2})^{-2}=$

$\frac{(c^{-2})^{-2}}{2^{-2}}=$

$\frac{c^4}{\frac{1}{2^2}}=$

$\frac{c^4}{\frac{1}{4}}=$

$4c^4$

5.

$\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=$

$\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=$

$\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=$

$\frac{(a^4bc^5)b^3}{(-1)(a^2)}=$

$\frac{-a^4b^4c^5}{a^2}=$

$-a^2b^4c^5$

3 0

3 years ago