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dlinn [17]
3 years ago
10

T=h+kg isolate g rearrange the formula above

Mathematics
1 answer:
Westkost [7]3 years ago
8 0

Answer:

\frac{\left(t-h\right)}{k}\\ = g

Step-by-step explanation:

Subtract the h

t = h+kg

-h  -h

you get:

t-h=kg

now divide the k from both sides

\frac{\left(t-h\right)}{k}\\ = \frac{\left(kg\right)}{k}

so,

\frac{\left(t-h\right)}{k}\\=g

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Solve the following equation: <br><br> x^3 -168 = [24(x-2)] / 2<br><br> PLEASE REPLY ASAP
Nataly_w [17]
Hi Larry

x^3 - 168 = [ 24(x-2)] / 2
x^3 - 168 = (24x - 48)/2
x^3 - 168 = 12x - 24
Subtract 12x - 24 to both sides
x^3 - 168 - (12x - 24) = 12xx - 24 - (12x - 24)
x^3 - 12x - 144 = 0
Now, factor the left sides
(x - 6)(x^2 + 6x + 24) = 0
Set factors equal to 0
x - 6 = 0 or x^2 + 6x + 24 = 0
x = 0 + 6 or x^2 + 6x + 24 - 0
x = 6
Answer : X = 6

Good luck !
7 0
3 years ago
Twice the difference between a number
Alecsey [184]

Answer:

Let x = the number

2(x-5) = 9

2x - 10 = 9

2x = 19

x = 9.5

Step-by-step explanation:

7 0
3 years ago
Window design. A triangular window on Ross camp's new scary boat has two sides that measure 87 centimeters 64 centimeter, respec
krek1111 [17]
Since the perimeter must not exceed 291.
Let the third side be x.

x + 87 + 64  < 291
x + 151 < 291.
x            < 291 -151.
x   <  140.      (First)

But for a triangle there is what is called the Triangle Inequality Theorem.  That given  the two sides of a tringle, the third side of the triangle must greater than the positive difference between the two sides and less than the sum of the two sides.

So for this case.   87 and 64.

x  > ( 87 - 64).    x  > 23.
x < (87 + 64)      x  <  151. Combine both inequalities.

       23  < x < 151  (second).

Combining First and second. Both must be satisfied.
So we have  a more accurate answer as:


23 < x < 140.      x is greater than 23 and x is less than 140.

x  could be    24, 25, 26, 27, ......, 139.  cm.

I hope this helps.
3 0
3 years ago
Read 2 more answers
When the expression $-2x^2-20x-53$ is written in the form $a(x+d)^2+e$, where $a$, $d$, and $e$ are constants, then what is the
Sladkaya [172]
We start with 2 x^{2} -20x-53 and wish to write it as a(x+d) ^{2} +e

First, pull 2 out from the first two terms: 2( x^{2} -10x)-53

Let’s look at what is in parenthesis. In the final form this needs to be a perfect square. Right now we have x^{2} -10x and we can obtain -10x by adding -5x and -5x. That is, we can build the following perfect square: x^{2} -10x+25=(x-5) ^{2}

The “problem” with what we just did is that we added to what was given. Let’s put the expression together. We have 2( x-5) ^{2}-53 and when we multiply that out it does not give us what we started with. It gives us 2 x^{2} -20x+50-53=2 x^{2} -20x-3

So you see our expression is not right. It should have a -53 but instead has a -3. So to correct it we need to subtract another 50.

We do this as follows: 2(x-5) ^{2}-53-50 which gives us the final expression we seek:

2(x-5) ^{2}-103

If you multiply this out you will get the exact expression we were given. This means that:
a = 2
d = -5
e =  -103

We are asked for the sum of a, d and e which is 2 + (-5) + (-103) = -106


7 0
3 years ago
If x + 10 = 70, what is x?<br> Α. 40<br> B. 50<br> C. 60<br> D. 70<br> E. 80
Helen [10]

Answer:

C.60

x + 10 = 70

x = 70 -10

x =60

4 0
2 years ago
Read 2 more answers
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