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pishuonlain [190]

2 years ago

9

Question 1 of 10 Solve 10x+16 ≥ 6x+ 20. O A. x29 OB. x≤ 1 OC. x2 1 OD. x≤ 9

1 answer:

Alexxandr [17]2 years ago

6 0

Answer: I think B. Explain:

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Find the slope of a line whose equation is 3X+6Y=9

Lisa [10]

To find the slope of the above equation, it is easiest to put it into slope-intercept form, y=mx + b, where the variable m represents the slope. To do this, we must isolate the variable y on the left side of the equation by using the reverse order of operations. First, we should subtract 3x from both sides of the equation.

3x + 6y = 9

6y = -3x + 9

Next, we should divide both sides of the equation by 6 to undo the coefficent of 6 on the variable y.

y = -1/2x + 3/2

Therefore, the slope of the line is -1/2 (the coefficient of the variable x in slope-intercept form).

Hope this helps!

8 0

3 years ago

Help I will mark you brainly

Ludmilka [50]

Answer:

<em>Hi</em><em> </em><em>there</em><em>!</em><em>!</em>

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<em><u>I</u></em><em><u> </u></em><em><u>hope</u></em><em><u> </u></em><em><u>it will</u></em><em><u> </u></em><em><u>help u</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

5 0

3 years ago

Read 2 more answers

1. 4x– 15 = 17 - 4x+10x<br> Help?

Nadya [2.5K]

Answer:

x=-16

Step-by-step explanation:

4x– 15 = 17 - 4x+10x

4x+4x-10x = 17+15

-2x=32

-x=16

x=-16

7 0

3 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago

Math question down below

Andrei [34K]

Eq. of given line is x = -2 》x + 2 = 0

6 0

3 years ago