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3 years ago

Which graph represents the function y=3x-5

Mathematics

2 answers:

elixir [45]3 years ago

7 0

Start at 5 on x line, then go 3 up one over and that’s how you find it

Elanso [62]3 years ago

6 0

Here is the answer for y=3x-5

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Hisako sells dolls at her doll store. If she sells a doll for $35, and there is 6% sales tax, what is the tax amount a customer

Sloan [31]

Given:

The price of a doll = $35

Sales tax = 6%

To find:

The tax amount that a customer will pay on one doll.

Solution:

The price of doll is $35 and the sales tax is 6%, so the tax amount on a doll is 6% of 35.

$\text{Tax amount}=35\times \dfrac{6}{100}$

$\text{Tax amount}=\dfrac{210}{100}$

$\text{Tax amount}=\dfrac{21}{10}$

$\text{Tax amount}=2.1$

Therefore, the tax amount that a customer will pay on one doll is $2.1.

7 0

2 years ago

How do I do this problem?

DedPeter [7]

If it's an isosceles triangle, the answer should be 87*42yd. In a right triangle, the area is (height*base)/2. if the height perfectly divides the triangle in half, there's no need to divide h*b by 2.

5 0

3 years ago

Alexis buys groceries using her credit card her card has an APR of 19.99% and she must pay at least 5% of the balance at the end

avanturin [10]

Answer:

The answer is that she would pay $65.56 in finance charges at the end of the month.

Step-by-step explanation:

Given: APR = 19.99%

Carry Over Balance: $398.97

The APR or Annual Percentage Rate, is calculated daily. You will need to get the daily periodic rate, or DPR, so divide the APR by 365:

19.99% = .1999

.1999 / 365 = .005477 (This is the Approximate DPR, rounded up to .005477)

To get the finance charge, multiply the average daily balance by the DPR and then by 30 days:

398.97 * .005477 * 30 = $65.56 finance charge for this carry over balance, at the end of the month. This assumes that the balance is the average daily balance.

Hope this helps!! Have a great day!

7 0

3 years ago

A rectangular garden must have a perimeter of 150 feet and an area of at least 1200 square feet. Describe the possible lengths o

kipiarov [429]

Answer: 51.86 and 23.14

Step-by-step explanation:

The rectangular garden must have a perimeter of 150 feet

Perimeter of rectangular garden =

2l + 2w=150 -----------1

The rectangular garden must have a perimeter of 150 feet and an area of at least 1200 square feet.

Area of rectangular garden =

l×w= 1200 feet^2 -----------2

From equation 2, l=1200/w

Put l=1200/w in equation 1

2× 1200/w + 2w = 150

(2400/w) +2w = 150

(2400+2w^2)/w =150

2400+2w^2= 150w

2w^2- 150w+2400=0

Using the general formula

w = [-b+-√(b^2-4ac)]/2a

a = 2, b =-150, c=2400

w =[--150+/-√(-150^2-4×2×2400)]/2×2

=[150+/-√(22500-19200)]/4

=[150+/-√3300)]/4

=(150+57.45)/4 or (150-57.45)/4

w= 207.45/4 or 92.55/4

w= 51.86 or w= 23.14

l = 1200/51.86 or l= 1200/23.14

l = 23.14. or l= 51.86

For an area of at least 1200ft^2

The dimensions are 51.86 and 23.14

4 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

2 years ago