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3 years ago

ILL GIVE BRAINLIEST TO WHOE ER ANSWERS

Mathematics

1 answer:

statuscvo [17]3 years ago

3 0

Answer:

I would say the answer would be a

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peter owns a lawn mowing service.For every 3 hours of lawn mowing,peter charges $28.80.Write an equation that models the relatio

Daniel [21]

$\$ 28.80\div 3=\$ 9.6$ - charge per hour

$\boxed{y=9.6h}$

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3 years ago

If the largest angle of a triangle is 120° and it is included between sides of 1.5 and 0.5, then (to the nearest tenth) the larg

12345 [234]

Answer:

1.8 units.

Step-by-step explanation:

The questions which involve calculating the angles and the sides of a triangle either require the sine rule or the cosine rule. In this question, the two sides that are given are adjacent to each other and the given angle is the included angle. This means that the angle is formed by the intersection of the two lines. Therefore, cosine rule will be used to calculate the length of the largest side of the triangle. The cosine rule is:

b^2 = a^2 + c^2 - 2*a*c*cos(B).

The question specifies that a=0.5, B=120°, and c=1.5. Plugging in the values:

b^2 = 0.5^2 + 1.5^2 - 2(0.5)(1.5)*cos(120°).

Simplifying gives:

b^2 = 3.25.

Taking square root on the both sides gives b = 1.8 (rounded to the nearest tenth).

This means that the length of the third side is 1.8 units!!!

3 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

Eric reflected parallelogram ABCD across the x-axis. If angle A is 125° and angle B is 55°, what is the degree measurement of an

Sauron [17]

The measures of the angles don't change when you translate a figure, because the entire figure is moving as a whole. Imagine having a paper parallelogram, moving it around and flipping it over. Not even dilations would change these angles (for reasons that can be pretty easily visualed but not really proven until geometry)

$m\angle A'=125\°$

7 0

3 years ago

ANSWER THIS ASAP NO FILE AT ATT!! IM AT 69 I NEED A 70 JUST GET IT RIGHT

Darya [45]

Answer:

Step-by-step explanation:

A shows a counterclockwise 90° rotation, so it is the correct answer.

B shows a reflection.

C shows a translation.

8 0

2 years ago