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Anuta_ua [19.1K]
3 years ago
12

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

<u>4)</u><u> LCM = 819</u>

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M∠1 = 80 – x; and
Burka [1]

Answer:

10

Step-by-step explanation:

<em>m∠</em>2 and <em>m∠</em>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.

6 0
2 years ago
18 = 3(3x - 6)<br> Please answer with solving steps thanks!
S_A_V [24]

Answer:

jayfeather friend me

Step-by-step explanation:

1

8

=

3

(

3

−

6

)

1

8

=

9

−

1

8

2

Add

1

8

18

18

to both sides of the equation

3

Simplify

4

Divide both sides of the equation by the same term

5

Simplify

Solution

=

4

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
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