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notsponge [240]
3 years ago
9

The ratio of girls to boys in jame’s chorus is 2 to 4. If there are 25 girls in her chorus, how many boys are there ?

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
3 0

Answer:

There are 50 boys

Step-by-step explanation:

girls to boys - 2:4

so 25x2=50

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Expand &amp; simplify <br> (2x+6)(x−9)
Alex777 [14]

Answer:

2x^2-12x-54

Step-by-step explanation:

First 2x times x along with 2x times -9

Second 6 times x  along with 6 times -9

2x^2-18x+6x-54

=2x^2-12x-54

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2 years ago
Last year 190 student attended the spring dance and the total cost was 1250. Two years ago 175 people attended and the total cos
kondor19780726 [428]
Unit cost of a ticket = Income from ticket sales / number of tickets sold:

$1250
--------------- = $6.58 per ticket
190 tickets

Again:

$1175
--------------- = $6.71
175 tickets

While ticket prices do change (usually increase) from year to year, it's unusual to see such a situation here.

Don't have any guidelines by which to determine the "fixed cost of a ticket".

If we use the cost of a ticket of 2 years ago ($6.58/ticket), then the income from the sale of 225 tickets this year would be ($6.58/ticket)(225 tickets), or $1480.50.
6 0
3 years ago
You can have a negative number over a negative number for slope. *<br> True<br> False
snow_lady [41]

Answer:

True

Step-by-step explanation:

7 0
3 years ago
How would you solve h(x)=12/x if h(x)= -2
MaRussiya [10]
H=-6 is the answer after substituting
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Graph f(x) = (1/3)^x+1<br>​
e-lub [12.9K]

Answer:

The graph is attached below.

Step-by-step explanation:

Given the function

f\left(x\right)\:=\:\left(1/3\right)^x+1

Exponentiation function have a horizontal Asymptotes.

<u />

<u>Finding the horizontal Asymptotes</u>

<u />\mathrm{Exponential\:function\:of\:the\:form}\:f\left(x\right)\:=c\cdot \:n^{ax+b}+k\:<u />

<u />\mathrm{has\:a\:horizontal\:asymptote}\:y=k\:<u />

<u />k=1<u />

<u />\mathrm{The\:horizontal\:asymptote\:is:}<u />

<u />y=1<u />

<u />

<u>Finding y-intercepts</u>

<u />y\mathrm{-intercept\:is\:the\:point\:on\:the\:graph\:where\:}x=0<u />

<u />y=\left(\frac{1}{3}\right)^0+1<u />

<u />\mathrm{Apply\:rule}\:a^0=1,\:a\ne \:0<u />

<u />\left(\frac{1}{3}\right)^0=1<u />

<u />y=1+1<u />

<u />\mathrm{Add\:the\:numbers:}\:1+1=2<u />

<u />y=2<u />

\mathrm{Y\:Intercepts}:\:\left(0,\:2\right)

The graph is attached below.

4 0
3 years ago
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