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3 years ago

How do I solve systems by elimination

Mathematics

2 answers:

Otrada [13]3 years ago

6 0

Answer: The most frequent way you can use it is subtracting!

Step-by-step explanation:

Lorico [155]3 years ago

3 0

In order to use the elimination method, you have to create variables that have the same coefficient—then you can eliminate them. Multiply the top equation by 5. Now multiply the bottom equation by −3. Next add the equations, and solve for y.

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What is another name for 23 ten thousands?

kicyunya [14]

<span>230,000

</span>Hope I Helped You!!! :-)

Have A Good Day!!!

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2 years ago

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A rock is thrown from the top of a tall building.The distance in feet between the rock and the ground t seconds after thrown is

marusya05 [52]

Answer:

0.448 seconds

Step-by-step explanation:

d = -16t² -4t + 412

find t when d = 407

substituting d = 407 into the equation:

407 = -16t² -4t + 412 (subtract 407 from both sides)

-16t² -4t + 412 - 407 = 0

-16t² -4t + 5 = 0 (multiply both sides by -1)

16t² + 4t - 5 = 0

solving using your method of choice (i.e completing the square or using the quadratic equation), you will end up with

t = (-1/8) (1 + √21)= -0.70 seconds (not possible because time cannot be negative)

t = (-1/8) (1 -√21) = 0.448 seconds (answer)

4 0

3 years ago

Can someone help me plz

agasfer [191]

I hope this helps you

8 0

3 years ago

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If f(x)...............

alex41 [277]

Answer:

The inverse is 5x-6

Step-by-step explanation:

To find the inverse of a function

y = ( x+6) /5

Exchange x and y

x = ( y+6)/5

Solve for y

Multiply each side by 5

5x = (y+6)/5 *5

5x = y+6

Subtract 6 from each side

5x-6 = y+6-6

5x-6 =y

The inverse is 5x-6

4 0

1 year ago

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How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

Integration by parts will help here. Letting $u=\arctan x$ and $\mathrm dv=\mathrm dx$ , you end up with $\mathrm du=\dfrac{\mathrm dx}{1+x^2}$ and $v=x$ . Now

$\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du$
$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$

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2 years ago