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3 years ago

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 a) 17 (b) 71 c) 7 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. (−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 y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 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 x 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 y = 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]

ST/VT = UT/WT6+14/14 = (x+4)+21/21X= 20*21/14 - 21x=420/14-25X=30-25x=5

3 0

3 years ago

Read 2 more answers

Is x+y=10 linear or non

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