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IRISSAK [1]
3 years ago
12

What is the value of x ?

Mathematics
1 answer:
statuscvo [17]3 years ago
8 0
Here, Sin Ф = P / H
Sin 31 = x / 430
0.515 = x /430 
x = 430 * 0.515
x = 221.45

In short, Your Answer would be 221.45

Hope this helps!
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Find the value of x - if x^3-7x² - x = 0<br>a) 17 (b) 71 c) 7<br>d) 1​
S_A_V [24]

Answer:

x = 0

Step-by-step explanation:

x^3-7x^2-x =0\\

Take out common factor

x(x^2-7x-1) = 0

The factor inside the parenthesis cannot be factored

Since there is a common factor that we pulled out of the original equation,

set that x equal to zero and solve

x = 0\\ : Therefore x = 0

6 0
3 years ago
Simplify the expression.<br><br> (−2g5h2)5
vampirchik [111]
I am sure you are trying to type

( - 2 {g}^{5} {h}^{2} {)}^{5}
If that is it, then you should recall the law of exponents which says,

({a}^{m} {)}^{n} = {a}^{mn}
Let us share the exponent for each of them.

( - 2 {g}^{5} {h}^{2} {)}^{5} = ( { - 2)}^{5} ( { {g}^{5}) }^{5} ( { {h}^{2}) }^{5}
We now apply the law,

( - 2 {g}^{5} {h}^{2} {)}^{5} = ( { - 2)}^{5} ( { {g}^{5 \: \times 5}) }^{} ( { {h}^{2 \times 5}) }^{}
We now simplify to obtain,

( - 2 {g}^{5} {h}^{2} {)}^{5} = -32 {g}^{25} {h}^{10}

If that is not the expression let me know so I will know how to help you
4 0
3 years ago
The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to
Firdavs [7]

Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:

<em>y</em> = -1 - <em>x</em>  ==>  <em>x</em> = -1 - <em>y</em>

<em>y</em> = <em>x</em> - 1  ==>  <em>x</em> = 1 + <em>y</em>

So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.

This means the area is given by the integral,

\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy

The integral with respect to <em>x</em> is trivial:

\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy

For the remaining integral, integrate term-by-term to get

\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0
3 years ago
Solve for x. Show the equation you used and your work.
kirza4 [7]
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3 0
3 years ago
Read 2 more answers
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fiasKO [112]

Answer:

Not

Step-by-step explanation:

In this case, the degree of variable y is 1 , the degrees of the variables in the equation violate the linear equation definition, which means that the equation is not a linear equation.

6 0
3 years ago
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