All categories

3 years ago

Plz help ill give you brainlist

Mathematics

1 answer:

tamaranim1 [39]3 years ago

5 0

Hey there!

To find the markup percentage, you must use your formula for percent markup:
profit/cost*100

Because your profit is found by subtracting 13-10, 3, your profit would be 3. You also know that your cost is 10.

Now, plug in your values:
3/10*100

And simplify:
3/10*100
0.3*100
=30

Therefore, your percent markup would be about 30%.

Hope this helps and have a nice day!

You might be interested in

The midpoint between 94.68 and 49.67

Brilliant_brown [7]

Answer:

72.175

Step-by-step explanation:

The first step is subtract 94.68 by 49.67 which would be 45.01

The next step would be to use 45.01 and divide by two

The last step would be subtract 94.68 by 22.505 and you would get 72.175

Therefore your answer would be 72.175

8 0

3 years ago

Pumpkins are being sold for $1.30 at the local farmers market. How much would a pumpkin be weighing 6lb 9oz

True [87]

Answer:

$8.53

Step-by-step explanation:

The first thing you do is convert the oz into pounds. There is 16 oz in a pound so our fraction is 9/16 which is 0.5625 . Then adding to the pounds it becomes 6.5625 pounds. Then multiply by 1.3.

8 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

Which direction would the graph of y=|x+5|-1 open? a. down b.up c.right d.left

amm1812

The answer fam is.......... "b.up"

3 0

3 years ago

Please help me on this

Anika [276]

The answer is: [A]:
___________________________________________________________
"According to the rules, how many swimmers are on a team?"
___________________________________________________________

8 0

3 years ago