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rosijanka [135]
3 years ago
10

Pete is saving money to buy a used car

Mathematics
1 answer:
Arlecino [84]3 years ago
7 0

Answer:

450 per month

Step-by-step explanation:

5000-1400=3600

3600/8=450

Since Peter already has 1400 dollars saved you can subtract that form the total amount and since he is trying to get the rest in eight months you just divide the difference by eight.

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URGENT! Describe step by step how to find the following equations.
yuradex [85]

Answer:

1. x = 45°

2. x = 330°

3. x = 105°

4. x = 30°

5. x = 60°

6. x = 30° or x = -45° (315°) or x = 45°

Step-by-step explanation:

1. For

cos x = sin x

∴ sin x/(cos x) = 1 = tan x

Hence, x = tan⁻¹1 = 45°

2. For

csc(x) + 3 = 1

∴ csc(x) = 1 - 3 = -2

Which gives sin x = -1/2

∴ x = sin⁻¹(-1/2) = -30° = 360 +(-30) = 330 °

3. For

cot(3x) = -1

∵ cot(3x) = 1/(tan(3x))

Hence, 1/(tan(3x)) = -1

∴ tan(3x) = -1

3·x = tan⁻¹(-1) = -45° = 360 + (-45) = 315°

Which gives, x = -45/3 = -15° or x = 105°

4. For

2·sin²(x) + 3·sin(x) = 2

We put sin(x) = y to get

2·y² + 3·y = 2 or 2·y² + 3·y - 2 =0

Factorizing gives

(2·y -1)(y+2) =0

∴ y = 1/2 or y = -2

That is, sin(x) = 1/2 or sin(x) = -2

Hence, x = sin⁻¹(1/2) = 30° or x = sin⁻¹(-2) = (-π/2 + 1.3·i)

∴ x = 30°

5. For

4cos²(x) = 3 we have;

cos²(x) = 3/4

cos(x) = √(3/4) = (√3)/2

∴ x = cos⁻¹((√3)/2) = 60°

6. For

4·sin³(x) + 1 = 2·sin²(x) + 2·sin(x)

We put sin(x) = y to get;

4·y³ + 1 = 2·y² + 2·y which gives;

4·y³ + 1 - (2·y² + 2·y) = 0 or 4·y³ -2·y² - 2·y + 1 = 0

Factorizing gives;

\frac{(2x-1)(2x+\sqrt{2}) )(2x-\sqrt{2})}{2} =0

Therefore, x = 1/2 or x = -(√2)/2 or (√2)/2

Therefore, sin(x) = 1/2 or -(√2)/2 or (√2)/2

That is x = sin⁻¹(1/2) = 30 or sin⁻¹(-(√2)/2) = -45 or sin⁻¹((√2)/2) = 45.

8 0
3 years ago
The equation below describes a parabola. If a is positive, which way does the
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B. Up

Because a positive ‘a’ value will make the parabola open upwards, and because the power is 2 (even) both sides will open up in a ‘U’ shape
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2 years ago
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A diagram of a rectangular pool with the diagonal of 50 meters is shown below. Connected to the pool is a square shaped kid’s po
NeTakaya
The pool is 40 since half of the other rectangle is 50 which the 2 halfs equal 100 and we already know the top side measurement which is 40 and 2 sides of 40 equal 80 and there’s 20 left so this concludes that’s the sides are 10 and square sides are congruent to each other which concludes that’s the AREA is 40
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Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20
Mice21 [21]

Power and chain rule (where the power rule kicks in because \sqrt x=x^{1/2}):

\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'

Simplify the leading term as

\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}

Quotient rule:

\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}

Chain rule:

(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)

(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)

Put everything together and simplify:

\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}

=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}

=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}

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