The answer of this problem please

Joel used a half gallon of paint to cover 84 square feet of wall. He has 932 square feet of wall to paint. How many gallons of p

DiKsa [7]

Answer:

6 Gallons of Paint

Step-by-step explanation:

If you use a half gallon of paint to cover 84 square feet, how much could you paint with a full gallon? Lets figure it out:

$\frac{1}{2} = 84$

multiply both sides by 2 so that the 1/2 gallon is now 1 gallon:

$\frac{1}{2} * 2 = 84 * 2$

1 gallon = 168 square feet

now, how many gallons does it take to paint 932 square feet?

Divide 932 by 168:

$\frac{932}{168} = 5.548$

Just like you cant buy 2.374 cheeseburgers, you cant buy 5.548 gallons of something so we round up and will have some paint left over once the job is done. So 6 gallons is the answer

4 0

3 years ago

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HELP ME GRAPH THIS EQUATION. I'LL GIVE BRAINLIEST.

scoray [572]

The y-intercept is at 1, from there plot the next point 1 down and three to the right (0,3), and repeat until the graph is done

6 0

3 years ago

catering service offers 8 appetizers, 11 main courses, and 7 desserts. A banquet committee is to select 7 appetizers, 8 main

guapka [62]

Answer: The required number of ways is 46200.

Step-by-step explanation: Given that a catering service offers 8 appetizers, 11 main courses, and 7 desserts.

A banquet committee is to select 7 appetizers, 8 main courses, and 4 desserts.

We are to find the number of ways in which this can be done.

We know that

From n different things, we can choose r things at a time in $^nC_r$ ways.

So,

the number of ways in which 7 appetizers can be chosen from 8 appetizers is

$n_1=^8C_7=\dfrac{8!}{7!(8-7)!}=\dfrac{8\times7!}{7!\times1}=8,$

the number of ways in which 8 main courses can be chosen from 11 main courses is

$n_2=^{11}C_8=\dfrac{11!}{8!(11-8)!}=\dfrac{11\times10\times9\times8!}{8!\times3\times2\times1}=165$

and the number of ways in which 4 desserts can be chosen from 7 desserts is

$n_3=^7C_4=\dfrac{7!}{4!(7-4)!}=\dfrac{7\times6\times5\times4!}{4!\times3\times2\times1}=35.$

Therefore, the number of ways in which the banquet committee is to select 7 appetizers, 8 main courses, and 4 desserts is given by

$n=n_1\times n_2\times n_3=8\times165\times35=46200.$

Thus, the required number of ways is 46200.

7 0

4 years ago

Help pls - 20pts<br><br> 2020202022020

Gala2k [10]

Your answer would be:
f(8)= 68,594
f(11)= 91,299
f(2)= 38,720
f(17)= 161,743

4 0

3 years ago

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PLEASE HELP!!

Aloiza [94]

Answer:

From the given graph:

the coordinates of triangle RST are;

R= (2, 1),

S= (2,-2),

T= (-1,-2)

Given: Scale factor = $\frac{5}{3}$ and center of dilation at (2,2)

The mapping rule for the dilation applied to the triangle as shown below:

$(x,y) \rightarrow (\frac{5}{3}(x-2)+2 , \frac{5}{3}(y-2)+2 )$ ; where k represents the scale factor i.e, $k=\frac{5}{3}$ or we can write it as ;

For R=(2, 1)

The image R' = $(\frac{5}{3}(2-2)+2 , \frac{5}{3}(1-2)+2 )$

⇒ R'= $(2, \frac{1}{3})$

Similarly for S= (2, -2) and T= (-1,-2)

therefore, the image of S'= $(\frac{5}{3}(2-2)+2 , \frac{5}{3}(-2-2)+2 )$

⇒ S'= $(2, \frac{-14}{3})$

The image of T' = $(\frac{5}{3}(-1-2)+2 , \frac{5}{3}(-2-2)+2 )$

⇒T' = $(-3, \frac{-14}{3})$

Now, labelling the image of triangle R'S'T' as shown in the figure given below

5 0

3 years ago

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