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3 years ago

NEED HELP ASAP PLEASE WITH EXPLANATION PLEASE

Mathematics

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer:

(1) 20%

Step-by-step explanation:

→ Find the difference in price

200 - 160 = 40

→ Divide difference by original price

40 ÷ 200 = 0.2

→ Multiply the answer by 100

0.2 × 100 = 20%

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here a subtraction problem that has been solved incorrectly for 44-16=28 choose the addition sentence that check this subtractio

Mazyrski [523]

Answer:

28+16=44

Step-by-step explanation:

point of correction it has been solved correctly

when subtracting 44 - 16 it gives you 28

=> 44 - 16

=> 28

SO IN ORDER TO CHECK IF THE PROBLEM WAS SOLVE CORRECTLY

you will minus 28 from one of the numbers before the equal sign

the correct option is 28 + 16 equals to 44

=> 28+16=44

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2 years ago

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3 years ago

Solve for x and y. Round answers to the nearest tenth.

kifflom [539]

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3 years ago

The power generated by an electrical circuit (in watts) as a function of its current ccc (in amperes) is modeled by: P(c)=-12c^2

g100num [7]

Answer:

5 amperes will produce the maximum power of 300 watts.

Step-by-step explanation:

The general form of a quadratic function presents the function in the form

$f(x)=ax^2+bx+c$

The vertex of a quadratic function is the highest or lowest point, also known as the maximum or minimum of a quadratic function.

We can define the vertex by doing the following:

Identify a, b, and c
Find, the x-coordinate of the vertex, by substituting a and b into

$x-coordinate =-\frac{b}{2a}$

Find, the y-coordinate of the vertex, by evaluating

$y-coordinate =f(x-coordinate )=f(-\frac{b}{2a} )$

We know that the power generated by an electrical circuit is modeled by

$P(c)=-12c^{2}+120c$

This function is a quadratic function.

To find the current that produce the maximum power you must

Identify a and b

a = -12 and b = 120

Find, the maximum current of the vertex, by substituting a and b into

$maximum-current =-\frac{b}{2a}$

$maximum-current =-\frac{120}{2(-12)}\\\\maximum-current = 5$

Find, the maximum-power, by evaluating

$maximum-power =P(maximum-current)=P(-\frac{b}{2a} )$

$P(5)=-12(5)^{2}+120(5)=300$

5 amperes will produce the maximum power of 300 watts.

We can check our work with the graph of the function $P(c)=-12c^{2}+120c$ and see that the maximum is (5, 300).

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4 years ago