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Svetllana [295]
3 years ago
13

PLEASE HELPPPP I WILL MARK YOU AS BRAINLIEST!!!!!

Mathematics
1 answer:
statuscvo [17]3 years ago
6 0

Answer:

Let a be the cost of each apple.

Let p be the cost of each pear.

<u>Q</u><u>1</u>

Cost of 12 apples = 12a

<u>Q</u><u>2</u>

Cost of 4 pears = 4p

<u>Q</u><u>3</u>

TOTAL COST = 12a + 4p

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1. Simplify this expression: 13+ (-12) - (-5) = ?<br> O A.-30<br> O B.6<br> O C.30<br> OD. -6
Virty [35]

Answer:

B.6

Step-by-step explanation:

When we add and subtract, we do it from left to right

13+ (-12) - (-5)

13+-12

13-12 = 1

1 - -5

A negative negative is a positive

1 +5

6

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Svet_ta [14]

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Step-by-step explanation:

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What is the place value of the number 9 in 328.095?
brilliants [131]

Answer:

The tenths place

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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
Suppose the true proportion of voters in the county who support a restaurant tax is 0.54. Consider the sampling distribution for
Eva8 [605]

Answer:

The correct answer is:

(a) 0.54

(b) 0.0385

Step-by-step explanation:

Given:

Restaurant tax,

p = 0.54

Sample size,

n = 168

Now,

(a)

The mean will be:

⇒ μ \hat{p}= p

         =0.54

(b)

The standard error will be:

\sigma \hat{p} = \sqrt{[\frac{p(1-p)}{n} ]}

    = \sqrt{[\frac{(0.54\times 0.46)}{168} ]}

    = \sqrt{[\frac{(0.2484)}{168} ]}

    = 0.0385

6 0
2 years ago
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