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Airida [17]
3 years ago
6

Suppose that a car normally sells for $16,500 and is now on sale for $14,000. What is the percent of discount?

Mathematics
1 answer:
MrMuchimi3 years ago
8 0
The price has decreased by 16500 - 14000 = 2500

To find percentage increase/decrease, we divide the value difference by the original value then multiply the answer by 100

Percentage decrease = [2500 ÷ 16500] ×100 = 15.2%
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KATRIN_1 [288]

Answer:

A. 3/2

Step-by-step explanation:

When two exponents have the same base, you can cancel the bases and set the exponents equal to each other

3x-2=x+1

3x=x+3

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2 years ago
Read 2 more answers
Find the direction angle of 3i+1j.<br><br> Please help me. Thank you
Alekssandra [29.7K]

Answer:

Correct answer:  ∝ = 18.43°

Step-by-step explanation:

Given coordinates represent one vector in x-y plane, let be v.

v = 3i +1j = xi + yj => x = 3 and y = 1

The angle that this vector builds with the positive direction to the x axis will be found using the tangent.

∝ = tan⁻¹ y/x = tan⁻¹ 1/3 = tan⁻¹ 0.333 = 18.43°

∝ = 18.43°

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4 0
3 years ago
A ski-lodge casts a 3-foot long shadow. at the same time, a 3.5 foot ski pole sticking vertically out of the snow casts a 0.5 fo
Alisiya [41]

Answer:

The lodge is 21 feet tall.

Step-by-step explanation:

Ski Lodge:

object/shadow

x/3

Ski Pole:

object/shadow

3.5/0.5

Solve for x:

x/3 = 3.5/0.5

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2 years ago
Use the substitution of x=e^{t} to transform the given Cauchy-Euler differential equation to a differential equation with consta
kherson [118]

By the chain rule,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dt}=x\dfrac{\mathrm dy}{\mathrm dx}

which follows from x=e^t\implies t=\ln x\implies\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x.

\dfrac{\mathrm dy}{\mathrm dt} is then a function of x; denote this function by f(x). Then by the product rule,

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac1x\dfrac{\mathrm df}{\mathrm dx}

and by the chain rule,

\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dt^2}

so that

\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}=x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}

Then the ODE in terms of t is

\dfrac{\mathrm d^2y}{\mathrm dt^2}+8\dfrac{\mathrm dy}{\mathrm dt}-20y=0

The characteristic equation

r^2+8r-20=(r+10)(r-2)=0

has two roots at r=-10 and r=2, so the characteristic solution is

y_c(t)=C_1e^{-10t}+C_2e^{2t}

Solving in terms of x gives

y_c(x)=C_1e^{-10\ln x}+C_2e^{2\ln x}\implies\boxed{y_c(x)=C_1x^{-10}+C_2x^2}

4 0
3 years ago
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