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2 years ago

A sweater that normally costs $32.00 is on sale for $24.00. What is the percent

Mathematics

1 answer:

Vlad1618 [11]2 years ago

6 0

Answer:

25 %

Step-by-step explanation:

Percentage is given by :

$\%=\dfrac{\text{original value-new value}}{\text{origial value}}\times 100$

Original cost of sweater is $32 and sale cost is $24. So,

$\%=\dfrac{32-24}{32}\times 100\\\\=25\%$

Hence, 25 % is the markdown on the sweater.

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2 years ago

Write three equations whose solution is x=3.5

My name is Ann [436]

Answer:

1. 2x + 23 = 30

2. 3x + 19.5 = 30

3. (10x - 20)/ 5 = 3

Explanation:

Work backward:

1. 2x + 23 = 30

Start:

x = 3.5

Multiply the equation by 2:

2x = 2 (3.5)
2x = 7

Subtract 30 from both sides

2x - 30 = 7 - 30
2x - 30 = -23

Add 23 and 30 to both sides:

2x + 23 = 30

2. 3x + 19.5 = 30

Start:

x = 3.5

Add 6.5 to both sides:

x + 6.5 = 10

Multiply by 3:

3(x + 6.5) = 30

Distributive property:

3x + 19.5 = 30

3. (10x - 20)/ 5 = 3

Start:

x=3.5

Multiply by 10:

10x = 35

Subtract 20:

10x - 20 = 15

Divide by 5:

(10x - 20)/ 5 = 15/5

(10x - 20) / 5 = 3

3 0

3 years ago

1.) What is the maximum number of zeros for the following function<br> f(x) = x - 2x2 - 11x + 12

Lesechka [4]

Answer:

-14x+12

Step-by-step explanation

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4 0

3 years ago

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$

$T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5$

###

Another way of doing this would be to solve for the coefficients $a,b,c,d,e,g$ in

$f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5$

by expanding the right hand side and matching up terms with the same power of $x$ .

5 0

2 years ago