In a group of 60 children, there are twice as many boys as girls. How many girls are there?

Answers

Answer 1

Answer: $g-girls\n2g-boys\n\ng+2g=60\n3g=60\ \ \ \ |divide\ both\ sides\ by\ 3\n\boxed{g=20}\leftarrow answer$

Answer 2

Answer: If there is twice as many boys then girls, then there are 20 girls because 3x20=60
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Are the fractions 1/3 and 3/9 equivalent?

Answers

yes it is equivalent if u multiply 1 times 3 u get 3 if u multiply 3 times 3 is 9 so they r equivalent hope this help:)

yes, the fractions are equivalent.
in order to solve this problem, you need both fractions to have the same denominators.
ask yourself if 3 can go into 9. it can.
you change the denominator of 3 to 9. but when you change the denominator, you also have to change the numerator. since 3 goes into 9 three times, you also multiply the numerator by 3.
1/3 times 3/3 equal 3/9. since 3/9 and 3/9 are the same, they are equivalent

ESTIMATION - The total bill for lunch was $12.72, of which $0.72 was sales tax. What was the sales tax rate rounded to the nearest whole percent?PLEASE HELP!

Answers

your answer is 13.44 and rounded to the whole is still going to be 13

Help with math please? (View image then tell me answer and tell me how you got it from step-by-step)

Answers

oh I remember these... just a lot of tedious work:/
but we are going to break the problem down by finding the volume of the half sphere and then the cylinder. After finding the measurements for both we just add them all together:)
first lets get the cylinder done
the formula for the can issss
$\pi r^(2) h$
so given that we know thatt
$\pi = 3.14$
r = 10
h = 16
so lets plug in our values
3.14 x 10² x 16
since they are all right by each other in the formula example we know that you multiply them. So lets solve that
3.14 x 10² x 16 = 5,024
So that means that the area of our cylinder is 5,024units³
Now before u go anywhere we still need to figure out the volume for the top:)
so the formula for our half sphere is
4/3 $\pi r$ ³
so lets find our value first
we know that 4/3 is just 4/3 soooo
4/3 = 4/3
$\pi = 3.14$
r = 10
so lets plug in our values
4/3 x 3.14 x 10³
now we just solve and 10³ = 1,000
4/3 x 3.14 x 1,000 = 4,186.66667units³
But since we are finding half of a sphere we are going to take that amount and divide it by two.
4,186.66667 ÷ 2 = 2,093.33334
So now we just add together our cylinder volume and half sphere
half sphere = 2,093.33334units³
cylinder = 5,024units³
2,093.33334 + 5,024 = 7,117.33334
soo the volume of our shape is 7,117.33334 or rounded out we could just say 7,117
answer = 7,117units³
hope this was helpful
and don't forget to
MARK ME BRIANLIEST! :D

A container holds 8 quarts of juice. How much is this in gallons? Use the table below. Include the correct unit in your answer. Unit Symbol Fact Volume fluid ounce floz cup c 1c = 8floz pint pt 1pt = 2c quart qt 1qt = 2pt gallon gal 1gal = 4qt

Answers

Since there are 4 quarts in a gallon, 8 quarts equals 2 gallons

4 maybe i am bad at math soi could be wrong

Plz help 20 points answers

Answers

Answer:A-3rd box

B-first box

C-first box

D-middle box

Step-by-step explanation:

The equation 7^2=a^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A, in astronomical units, AU. If planet Y is k times the mean distance from the sun as planet X, by what factor is the orbital period increased?

Answers

Answer:

The period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

Step-by-step explanation:

The equation $T ^ 2 = a ^ 3$ shows the relationship between the orbital period of a planet, T, and the average distance from the planet to the sun, A, in astronomical units, AU. If planet Y is k times the average distance from the sun as planet X, at what factor does the orbital period increase?

For the planet Y:

$T_y ^ 2 = a_y ^ 3$

For planet X:

$T_x ^ 2 = a_x ^ 3$

To know the factor of aumeto we compared $T_x$ with $T_y$

We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:

$a_y ^ 3 = (a_xk) ^ 3$

$(T_y ^ 2)/(T_x ^ 2)=(a_y ^ 3)/(a_x^ 3)\n\n(T_y ^ 2)/(T_x ^ 2)=((a_xk)^3)/(a_x ^ 3)\n\n(T_y^ 2)/(T_x^ 2)=\frac{k ^ {3}a_(x)^ 3}{a_(x)^ 3}\n\n(T_(y)^ 2)/(T_(x)^ 2)=k ^ 3\n\nT_(y)^ 2 = T_(x)^(2)k^(3)\n\nT_(y) =k^{(3)/(2)}T_x$

Finally the period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

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