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2 years ago

Help me please if i dont do my homework my mom will beat me

Mathematics

2 answers:

WINSTONCH [101]2 years ago

8 0

Answer:

i think its 3.5

Step-by-step explanation:

you divide 21 by 6 i think, and that should give you what its increasing by

Snezhnost [94]2 years ago

4 0

3.5 is the answer

I did the math and that is what i got.

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Please help 6th grade math i will give brainliest to answer with worked shown

Wewaii [24]

Answer:

Step-by-step explanation:

She drinks 4 ounces every 30 minutes.

4*2= 8 She drinks 8 ounces in an hour.

8*8=64 She drinks 64 ounces in 8 hours

8 0

2 years ago

F(x) = 3x + 6 please help quickly!!!!!!!

Georgia [21]

Given the function, f(x) = 3x + 6, we can solve for f(a), f(a + h) and $\frac{(f(a + h) - f(a)) }{h}$ by substituting their values into f(x) = 3x + 6. We will have the following:

$\mathbf{f(a) = 3a + 6}\\\\\mathbf{f(a + h) = 3a + 3h + 6}\\\\\mathbf{\frac{(f(a + h) - f(a)) }{h} = 6}$

Given:

f(x) = 3x + 6

We are told to find:

f(a)
f(a + h), and
$\frac{(f(a + h) - f(a)) }{h}$

1. Find f(a):

Substitute x = a into f(x) = 3x + 6

f(a) = 3(a) + 6

f(a) = 3a + 6

2. Find f(a + h):

Substitute x = a + h into f(x) = 3x + 6

f(a + h) = 3(a + h) + 6

f(a + h) = 3a + 3h + 6

3. Find $\frac{(f(a + h) - f(a)) }{h}$ :

Plug in the values of f(a + h) and f(a) into $\frac{(f(a + h) - f(a)) }{h}$

Thus:

$\frac{((3a + 3h + 6) - (3a + 6)) }{h}\\\\\frac{(3a + 3h + 6 - 3a - 6) }{h}\\\\$

Add like terms

$\frac{(3a - 3a + 3h + 6 - 6) }{h}\\\\= \frac{3h }{h}\\\\\mathbf{= 3}$

Therefore, given the function, f(x) = 3x + 6, we can solve for f(a), f(a + h) and $\frac{(f(a + h) - f(a)) }{h}$ by substituting their values into f(x) = 3x + 6. We will have the following:

$\mathbf{f(a) = 3a + 6}\\\\\mathbf{f(a + h) = 3a + 3h + 6}\\\\\mathbf{\frac{(f(a + h) - f(a)) }{h} = 6}$

Learn more here:

brainly.com/question/8161429

6 0

2 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6) 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,

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