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Alex17521 [72]
3 years ago
12

An animal hospital advertises a veterinarian’s starting salary as $50,000 for the first year, with increases of $1,500 each year

for the next nineteen years. The total salary is represented by the series shown. (first image)
Which is equivalent to the series? Check all that apply.

Mathematics
1 answer:
ryzh [129]3 years ago
6 0

Answer:

A B & D

Step-by-step explanation:

Edge 2021

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-Solve the proportion (round to the nearest tenth).<br><br> 1910=x12
zysi [14]

The value of x in the proportion is 15.8.

1910 = x12

1910 = 12x

divide both sides by  12

1910 / 12 = 12x / 12

x = 15.8333333333

x ≈ 15.8

read more: brainly.com/question/14026746?referrer=searchResults

5 0
3 years ago
How much would $125 invested at 8% interest compounded continuously be worth after 16 years? Round your answer to the nearest ce
lidiya [134]

Answer:

D.

Step-by-step explanation:

Just did it. hope it helps!

3 0
3 years ago
The gas tank in Sharon’s car can hold up to 16.5 gal of gas. About how many liters of gas can the tank hold? 1 L≈1.06 qt
My name is Ann [436]
Most people have trouble deciding on what conversion factor to use. If there you use one conversion factor, and you get something like this:

(gal^2)/qt

Then you know that you used the wrong conversion factor. ALWAYS remember to multiply and divide your units. If you don't, you will not know whether or not you got what you were looking for.

4 0
3 years ago
Quadrilateral EFGH has coordinates E(a, 2a), F(3a, a), G(2a, 0), and H(0, 0). Find the midpoint of line HG
expeople1 [14]
Hope it helps!
#MissionExam001

6 0
3 years ago
Read 2 more answers
Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you
Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}

With \pi, we should expect \cos x. If \tan x=\dfrac8{15}, then

\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}

Also,

\pi

so we should expect \tan\dfrac x2 and

\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4

You seem to be taking

\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}

but this is not an identity.

###

8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

\dfrac{\sqrt6-\sqrt2}4

###

9. Use the identity from (7). \dfrac{7\pi}{12} lies in the second quadrant, so

\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3

###

12.

\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}

=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}

=\dfrac{\cos x}{\sin x}

=\dfrac{\csc x}{\sec x}

###

13.

\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}

=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}

=\dfrac1{\cos x}

=\sec x

###

14.

\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)

=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}

=2

###

15.

\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}

=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}

=\cos2\theta

3 0
3 years ago
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