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3 years ago

-6x-192=4x+58 I don’t understand how to find the X IN THIS

Mathematics

1 answer:

iren2701 [21]3 years ago

4 0

Answer:

Step-by-step explanation:First you cross out 192 and add it to 58. Then you cross out 4x and subtract it from -6x which gets you -10x. Your equation should end up being -10x= 250 and your final answer will be -25

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Determined to find the slope<br>(1 7K), and (-3,5K)

Helga [31]

Answer: $\dfrac{K}{2}$ .

Step-by-step explanation:

As we know ,

The slope of a line that passes through $(x_1,y_1)$ and $(x_2,y_2)$ is given by :

$\dfrac{y_2-y_1}{x_2-x_1}$

The slope of a line that passes through (1 , 7K), and (-3,5K) =

$\dfrac{5K-7K}{-3-1}\\\\=\dfrac{-2K}{-4}\\\\=\dfrac{K}{2}$

Hence, the slope of the given line is $\dfrac{K}{2}$ .

8 0

3 years ago

V1 - 2 sin cos 0 = cos 0 - sin

kramer

$\sqrt{1 - 2sin θ \cosθ } = \sqrt{ {sin}^{2} θ + {cos}^{2}θ - 2 sinθcosθ } = \sqrt{ {(cosθ - sinθ) }^{2} } = cosθ - sinθ$

4 0

3 years ago

2. Simplify:<br> b. 3(2x + 4y)

Nutka1998 [239]

Answer:

(6x+12y)

Step-by-step explanation:

answer is helpful or not

3 0

3 years ago

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The family of solutions to the differential equation y ′ = −4xy3 is y = 1 √ C + 4x2 . Find the solution that satisfies the initi

slega [8]

Answer:

The correct option is 4

Step-by-step explanation:

The solution is given as

$y(x)=\frac{1}{\sqrt{C+4x^2}}$

Now for the initial condition the value of C is calculated as

$y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(-2)=\frac{1}{\sqrt{C+4(-2)^2}}\\4=\frac{1}{\sqrt{C+4(4)}}\\4=\frac{1}{\sqrt{C+16}}\\16=\frac{1}{C+16}\\C+16=\frac{1}{16}\\C=\frac{1}{16}-16$

So the solution is given as

$y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}$

Simplifying the equation as

$y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1-256+64x^2}{16}}}\\y(x)=\frac{\sqrt{16}}{\sqrt{{1-256+64x^2}}}\\y(x)=\frac{4}{\sqrt{{1+64(x^2-4)}}}$

So the correct option is 4

8 0

3 years ago

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The graph of an equation with a negative discriminant always has which characteristic?<br>

jasenka [17]

Answer:

a posative

Step-by-step explanation:

3 0

3 years ago