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Effectus [21]
3 years ago
6

The dimensions of a rectangular prism are shown in the diagram below. Write an expression that represents the volume of the rect

angular prism?
Hint: Aprism= Length x width x height

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
7 0

Answer:

V=(a^{4})(b^{6})\ units^3

Step-by-step explanation:

we know that

The volume of a rectangular prism is equal to

V=LWH

In this problem we have

L=ab^2\\W=ab^3\\H=a^2b

substitute

V=(ab^2)(ab^3)(a^2b)

Applying property of exponents

x^{m} x^{n}x^{r} =x^{m+n+r}

V=(a^{4})(b^{6})\ units^3

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either 37, 17, 20 or 24, 26, 51

4 0
2 years ago
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April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3
vlada-n [284]

Answer:

        Student                                                            Hours worked

             April.                                                                  7\frac{7}{8} \ hrs

        Debbie.                                                                   8\frac{1}{8}\ hrs

        Richard.                                                                   7\frac{19}{24}\ hrs

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = 5\frac{1}{4}\ hrs

5\frac{1}{4}\ hrs can be Rewritten as \frac{21}{4}\ hrs

Number of Hours Carl worked on Math project = \frac{21}{4}\ hrs

Number of Hours Sonia worked on Math project = 6\frac{1}{2}\ hrs

6\frac{1}{2}\ hrs can be rewritten as \frac{13}{2}\ hrs

Number of Hours Sonia worked on Math project = \frac{13}{2}\ hrs

Number of Hours Tony worked on Math project = 5\frac{2}{3}\ hrs

5\frac{2}{3}\ hrs can be rewritten as \frac{17}{3}\ hrs.

Number of Hours Tony worked on Math project = \frac{17}{3}\ hrs.

Now Given:

April worked 1\frac{1}{2} times as long on her math project as did Carl.

1\frac{1}{2}  can be Rewritten as \frac{3}{2}

Number of Hours April worked on math project = \frac{3}{2} \times Number of Hours Carl worked on Math project

Number of Hours April worked on math project = \frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs

Also Given:

Debbie worked 1\frac{1}{4} times as long as Sonia.

1\frac{1}{4}  can be Rewritten as \frac{5}{4}.

Number of Hours Debbie worked on math project = \frac{5}{4} \times Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = \frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs

Also Given:

Richard worked 1\frac{3}{8} times as long as tony.

1\frac{3}{8} can be Rewritten as \frac{11}{8}

Number of Hours Richard worked on math project = \frac{11}{8} \times Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = \frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs

Hence We will match each student with number of hours she worked.

        Student                                                            Hours worked

             April.                                                                  7\frac{7}{8} \ hrs

        Debbie.                                                                   8\frac{1}{8}\ hrs

        Richard.                                                                   7\frac{19}{24}\ hrs

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3 years ago
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Max spent $53 and now has no money left. He had $ before his purchase.
Yuri [45]
Can you restate it or rewrite it? It doesn't have enough for me to answer this.
8 0
3 years ago
Am I dealing with harmonic stuff (trig)?
pochemuha

You just need to solve for when y=0:

\dfrac{\cos8t-9\sin8t}4=0\implies\cos8t-9\sin8t=0

\implies\cos8t=9\sin8t

\implies\dfrac19=\dfrac{\sin8t}{\cos8t}=\tan8t

\implies8t=\tan^{-1}\dfrac19+n\pi

\implies t=\dfrac18\tan^{-1}\dfrac19+\dfrac{n\pi}8

where n is any integer. We only care about when 0\le t\le1, which happens for n\in\{0,1,2\}.

t=\dfrac18\tan^{-1}\dfrac19\approx0.01

t=\dfrac18\tan^{-1}\dfrac19+\dfrac\pi8\approx0.41

t=\dfrac18\tan^{-1}\dfrac19+\dfrac\pi4\approx0.80

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3 years ago
Negative four times a number plus nine is no more than the number minus twenty one
marysya [2.9K]
N is no more than six
8 0
3 years ago
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