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NeX [460]
3 years ago
15

What is the solution to this equation? 2x+3=x-4​

Mathematics
1 answer:
g100num [7]3 years ago
4 0

Answer:

x = -7

Step-by-step explanation:

2x+3 = x-4

2x-x = -4-3

x = -7

hope it helps ☺️

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If K is the midpoint of JL, JK = 8x+11 and KL = 14x-1, find JL.
gladu [14]

Step-by-step explanation:

Since, K is the midpoint of JL.

\therefore \: JK = KL \\ 8x + 11 = 14x - 1 \\ 11 + 1 = 14x - 8x \\ 12 = 6x \\  \frac{12}{6}  = x \\ 2 = x \\  \\  \because \: JL = JK + KL \\   \therefore \: JL = 8x + 11 + 14x - 1 \\ \therefore \: JL = 8x + 14x + 11 - 1 \\ \therefore \: JL = 22x + 10\\  \therefore \: JL = 22 \times 2 + 10..(plug \: x = 2) \\  \therefore \:JL = 44 + 10 \\  \huge \red{ \boxed{\therefore \:JL = 54}}

7 0
3 years ago
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6% of what number is 30
uranmaximum [27]
1.8 is 6 percent of 30
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How can 5/16be written as a decimal
juin [17]

Answer:

5/16 as a decimal is 0.3125.

5 0
2 years ago
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Asibi makes face covers. A customer has ordered five face covers. Asibi has 3/4 yard of fabric and each face cover requires 1/6
Volgvan

Answer:

1/12 yards

Step-by-step explanation:

5×1/6 yards = 5/6 yards

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3 0
3 years ago
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
6 0
3 years ago
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