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Sladkaya [172]
3 years ago
13

The price of a jacket was reduced from $80 to $64. By what percentage was the price of the jacket reduced?

Mathematics
1 answer:
aleksley [76]3 years ago
7 0

Answer:

The asnwer is 20%

Step-by-step explanation:

80 times .20 is 16, 80-16 is 64, hoipe i helped!

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A local little has a total of 60 players ,of whom 40%are right-handed how many are right handed
maks197457 [2]

Answer:

A

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Given that A, O &amp; B lie on a straight line segment, evaluate <br>x<br>.​
Afina-wow [57]

Answer:

Step-by-step explanation:

It is the same way:

       we know that

if A , O and B lie on a straight line segment

then

∡AOY+∡YOZ+∡ZOB=180°

so

3x+x+x=180-----> 5x=180--------> x=180/5--------> x=36°

the answer is

x=36°

<em>please mark me as brainliest</em>

4 0
3 years ago
:P ??????<br><br><br><br><br><br> “ItS ToO ShOrT”
Rus_ich [418]

Answer:

The answer is >.

Step-by-step explanation:

78 19/88 is bigger than 74/93 because 74/93 has way less whole numbers on its side.

4 0
3 years ago
Jasmine wants to have an equal quantity of forks and spoons, what is the least number of packages of forks jasmine should buy to
vlada-n [284]

Answer:

6 packages of forks

Step-by-step explanation:

If Jasmine wants to have an equal quantity of forks and spoons, we need to list the multiples of each quantity and determine the least common multiple (LCM).

Forks: 10, 20, 30, 40, 50, 60, 70, 80, 90

Spoons: 12, 24, 36, 48, 60, 72, 84, 96

The LCM in this example is 60. In order to have exactly 60 forks and 60 spoons, Jasmine will need to buy 6 packages of forks [60 ÷ 10 = 6] and 5 packages of spoons [60 ÷ 12 = 5].

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
3 years ago
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