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algol13
3 years ago
13

I have a hard time finding the x help

Mathematics
2 answers:
Ratling [72]3 years ago
8 0
The volume of a triangular prism is V = 1/2lwh. So, in order to find x, plug in the values you know and solve:
576 =  \frac{1}{2}*8*8*x
1152 = 64x
x = 18
Mamont248 [21]3 years ago
3 0
The formula for the volume of this shape is: length * width * height/2

Knowing this, we can set up this equation:

576= \frac{8*8*x}{2}

Now, we can solve for x.  I will start by multiplying the 8s and then divide both sides by two.

1152=64x

Next, we divide both sides by 64 to find x.

x = 18

So, x is equal to 18 cm.


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which of the following is a polynomial expression a. 2^x + x + 3 b. x^2 + 2x + 7 or c. 1/x + 1/x2 + 1/x3 ( btw the ^2 means it’s
NeTakaya

(b)

A polynomial has the form

a_{0}x^{n} ± a_{1}x^{n-1} + ....

the only expression fitting this description is : x² + 2x + 7


5 0
3 years ago
What is 57.2 times 56.4
Rainbow [258]

Answer:

3226.08

Step-by-step explanation:

57.2 x 56.4

6 0
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Chris owes $17,000 to a bank. The debt grows with 8% annual interest that compounds every year. How
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A = 17000 (1 + .08/1)^(1*1)

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3 years ago
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For the function f(x)=5x2−5x, evaluate and simplify.
Marianna [84]

9514 1404 393

Answer:

  10x -5 +5h

Step-by-step explanation:

Use the function definition with the given arguments.

  \dfrac{f(x+h)-f(x)}{h}=\dfrac{(5(x+h)^2-5(x+h))-(5x^2-5x)}{h}\\\\=\dfrac{(5(x^2+2hx+h^2)-5(x+h))-(5x^2-5x)}{h}\\\\=\dfrac{5x^2+10hx+5h^2-5x-5h-5x^2+5x}{h}=\dfrac{10hx-5h+5h^2}{h}\\\\=\boxed{10x-5+5h}

7 0
3 years ago
Find the vector projection of B onto A if A = 5i + 11j – 2k,B = 4i + 7k​
valkas [14]

Answer:

\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\

Step-by-step explanation:

Given A = 5i + 11j – 2k and B = 4i + 7k​, the vector projection of B unto a is expressed as proj_ab = \dfrac{b.a}{||a||^2} * a

b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)

note that i.i = j.j = k.k  =1

b.a = 5(4)+11(0)-2(7)

b.a = 20-14

b.a = 6

||a|| = √5²+11²+(-2)²

||a|| = √25+121+4

||a|| = √130

square both sides

||a||² = (√130)

||a||²  = 130

proj_ab = \dfrac{6}{130} * (5i+11j-2k)\\\\proj_ab = \frac{30}{130} i+\frac{11}{130} j-\frac{12}{130} k\\\\proj_ab = \frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\\\

<em>Hence the projection of b unto a is expressed as </em>\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\<em></em>

7 0
3 years ago
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