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masya89 [10]
4 years ago
5

Please help me , algebra math !!

Mathematics
1 answer:
Ksju [112]4 years ago
3 0
Check the picture below.

\bf A)\\\\
V(w)=(w+2)(w)\left( \cfrac{w+2}{4} \right)\implies V(w)=(w^2+2w)\left( \cfrac{w+2}{4} \right)
\\\\\\
V(w)=\cfrac{w^3+2w^2+2w^2+4w}{4}\implies V(w)=\cfrac{w^3+4w^2+4w}{4}
\\\\\\
B)\\\\
V(10)=\cfrac{(10)^3+4(10)^2+4(10)}{4}\implies V(10)=\cfrac{1000+400+40}{4}
\\\\\\
V(10)=360

what does it tell us?  hmmm when w = 10, h = 3 and l = 12, that just means is a short fat box like the one in the picture.

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PLEASE ANSWER ILL MAKE YOU BRAINLIEST 2 questions
GREYUIT [131]

Answer:

Wait what is the question the question you put in was

"PLEASE ANSWER ILL MAKE YOU BRAINLIEST 2 questions"

Step-by-step explanation:

4 0
3 years ago
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Look at screen shot worth 15 points
soldi70 [24.7K]

Answer:

it would be by 23. that's what I think

3 0
2 years ago
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Solve the below: <br>4x+3y=7 3x-2y=9<br><br>​
Ahat [919]
4x+3y=7
3x-2y=9

12x+9y=21
12x-8y=36
^^Multiply top equation by 3 and multiply bottom equation by 4

12x+9y=36
- 12x-8y=36

17y=-15
^^Then subtract those two new equations

17y/17= -15/17

Y= -0.88
^^Next divide both sides by 17

4x+3* -0.88=7

4x + (-2.64) =7
-(-2.64) -(-2.64)

4x = 9.64

4x/4=9.64/4
X=2.41

^^then choose on equation and put your new y in the y spot and solve.


X=2.41

Y=-0.88

Your point/coordinate is (2.41,-.88)
7 0
3 years ago
If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)
Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

\log_dabc=\log_da+\log_db+\log_dc

Then using the change of base formula, we can derive the relationship

\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}

This immediately tells us that

\log_dc=\dfrac1{\log_cd}=\dfrac12

Notice that none of a,b,c,d can be equal to 1. This is because

\log_1x=y\implies1^{\log_1x}=1^y\implies x=1

for any choice of y. This means we can safely do the following without worrying about division by 0.

\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}

so that

\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23

Similarly,

\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}

so that

\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34

So we end up with

\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}

###

Another way to do this:

\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}

\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}

\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12

Then

abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}

So we have

\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}

4 0
3 years ago
Graph y &gt; x^2 - 5. Click on the graph until the correct one appears.
kenny6666 [7]

y=x^2\\\\for\ x=\pm2\to y=(\pm2)^2=4\to(-2;\ 4);\ (2;\ 4)\\\\for\ x=\pm1\to y=(\pm1)^2=1\to (-1;\ 1);\ (1;\ 1)\\\\for\ x=0\to y=0^2=0\to (0;\ 0)

Shift the graph of the function y = x², 5 units down /look at the picture #1/.

y > x^2-5 /look at the picture #2/ - your answer

7 0
3 years ago
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