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3 years ago

Solve for xhelp please!!!

Mathematics

2 answers:

VashaNatasha [74]3 years ago

8 0

Answer:

x=1

Step-by-step explanation:

Genrish500 [490]3 years ago

4 0

Answer:

7/3

Step-by-step explanation:

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8k + 2m = 3m + k<br><br> Solve for k

mina [271]

Minus both sides by 2m
$8k = 3 m - 2m + k \\ 8k = m + k$
Minus both sides by k
$7k = m$
Divided both sides by 7
$k = \frac{m}{7}$

6 0

3 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago

X+y=13 and 1/2x+y=10

photoshop1234 [79]

X=6
y=7

6+7
= 13

1/2(6) + 7
= 3 +7
=10

6 0

3 years ago

Plz help me and thank you

zimovet [89]

Answer:

i think im late

Step-by-step explanation:

4 0

2 years ago

Write an equation of the line through (3,-5) and perpendicular to 3y=x-6

Alona [7]

3y = x - 6
y = 1/3x - 2.
slope her is 1/3. A perpendicular line will have anegative reciprocal slope. To find the negative reciprocal of a number, u flip the number and change the sign. So the slope we need os -3/1 or just -3.(see how I flipped the slope and changed the sign)

y = mx + b
slope(m) = -3
(3,-5)...x = 3 and y = -5
now we sub and find b, the y int
-5 = -3(3) + b
-5 = -9 + b
-5 + 9 = b
4 = b

so ur perpendicular equation is : y = -3x + 4

3 0

3 years ago

Solve for xhelp please!!!​

Solve for xhelp please!!!