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zhenek [66]
3 years ago
6

What is the total volume of this composite shape? Round your answer to the nearest tenth.

Mathematics
1 answer:
prisoha [69]3 years ago
8 0
The answer is 240in^2
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5/7 times 28 cause I don’t know
Trava [24]

5/7 times 28 is 20

hope this helps :')

6 0
3 years ago
Read 2 more answers
Sue’s Corner Market has a markup of 60% on bottled water. If the market sells a bottle of water for $2, find the original amount
denis-greek [22]

Answer:

1.25

Step-by-step explanation:

original price + markup = retail

The markup = original * markup percentage

original  + original * markup percent= retail

Factor out the original

original( 1+ markup percent) = retail

original ( 1 + 60%) = 2

Change to decimal form

original ( 1+ .60) = 2

original ( 1.60) = 2

Divide each side by 1.6

original = 2/1.6

original =1.25

5 0
4 years ago
Write an equation for when slope =-3 and y-intercept = 36
Elenna [48]

Answer:

-3x+36

Step-by-step explanation:

the slope is -3, so just add an x.

the y intercept is 36

put them a together and you get -3x+36

6 0
3 years ago
Read 2 more answers
Divide 28 into 2 groups so the ratio is 3 to 4
emmasim [6.3K]

3     4

3     4

3     4

3     4

Left side gets 12.

Right side gets 16.

12+16=28 in total.

When you're thinking of ratios, you have to think of rounds. Let's say I want to distribute apples... In Round 1, I'm going to give myself 1 apple and give you 2 apples. After Round 1, three apples have been distributed in total. Now, if I repeat the same pattern of distribution 1:2, after Round 2 I should have 2 apples whilst you have 4. After Round 2 a total of 6 apples would have been distributed.If the ratio is 3 to 4 - imagine there are 7 groups total, and one person gets 3 of the 7 and the other person gets 4 of the 7.

____________________________________________
If 28 is divided into 7 groups, each group has 4 cans.

One person gets 3 groups = 4*3 = 12 cans
The other gets 4 groups = 4*4* = 16 cans

This way is easier for me to do, but go with whatever works for you :)
8 0
4 years ago
I need help with this problem
lubasha [3.4K]

at x = 1 we have one tangent line and at x = 5 we have just another tangent line.

f(x)=3x^2-15x\implies \left. \cfrac{df}{dx}=6x-15 \right|_{x=1}\implies \stackrel{\stackrel{m}{\downarrow }}{-9}~\hfill \left. 6x-15\cfrac{}{} \right|_{x=5}\implies \stackrel{\stackrel{m}{\downarrow }}{15}

so we have the slopes, but what about the coordinates?

well, for the first one we know x = 1 and we also know f(x), let's use f(1) to get "y", and likewise we'll do the for the second one.

\stackrel{x=1}{f(1)}=3(1)^2-15(1)\implies f(1)=-12\qquad \qquad (\stackrel{x_1}{1}~~,~~\stackrel{y_1}{-12}) \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-12)}=\stackrel{m}{-9}(x-\stackrel{x_1}{1}) \\\\\\ y+12=-9x+9\implies y=-9x-3 \\\\[-0.35em] ~\dotfill

\stackrel{x=5}{f(5)}=3(5)^2-15(5)\implies f(5)=0\qquad (\stackrel{x_1}{5}~~,~~\stackrel{y_1}{0}) \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{15}(x-\stackrel{x_1}{5})\implies y=15x-75

3 0
3 years ago
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