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3 years ago

HELP !! I NEED SOMEONE TO DO THIS TO ME!!

Mathematics

1 answer:

yKpoI14uk [10]3 years ago

4 0

Answer:

The LCM of 63 and 91 is 819

Explanation:

1) Find the prime number of 63

63 = 3 × 3 × 7

2) Find the prime number of 91

91 = 7 × 13

3) Multiply each factor the greater number of times it occurs in steps or above to find the LCM:

LCM = 3 × 3 × 7 × 13

4) LCM = 819

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M∠1 = 80 – x; and

Burka [1]

Answer:

$10$

Step-by-step explanation:

m∠2 and m∠1 are congruent by the Corresponding Angles Theorem; we set both expressions equal to each other:

[80 - x]° = [90 - 2x]°

+ x° + x°

______________

80° = [90 - x]°

-90° - 90°

__________

−10° = [−x]°

$10° = x$

I am joyous to assist you anytime.

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2 years ago

18 = 3(3x - 6) Please answer with solving steps thanks!

S_A_V [24]

Answer:

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Step-by-step explanation:

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3 0

3 years ago

Can somebody help me with this math question?

kap26 [50]

Answer:Hk and IJ

Step-by-step explanation:

They are parallel to eachother, hope this helped!

<!> Brainliest is appreciated! <!>

6 0

3 years ago

Help me out!!

vekshin1

Answer:

~Exponential decay because the base is less than 1~

Step-by-step explanation:

I just took the test and got a 100% :) hope this helped! Good luck!

8 0

3 years ago

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

3 years ago