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max2010maxim [7]

3 years ago

7

A Discount clothing store is selling a $72 sport jacket for $24 off the regular price. What is the discount rate?

1 answer:

mr Goodwill [35]3 years ago

7 0

72-24=48

x% of 72=48

(/72)x/100 *72=48 (/72)

(*100 ) x/100=2/3 (*100)

x= 66.67%

100% -66.67%=33.33% discount

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Set F1 = 5N at 0 degrees, F2= 5N at 90 degrees, F3 = 5N at 270 degrees and run the simulation. Using trigonometry, what net forc

Mrac [35]

Answer: find the solution in the explanation

Step-by-step explanation:

Let's use resolution of forces by resolving into x - component and y- component.

X - component.

Sum of forces = F1 - F3 - F4cos 15

Sum of forces = 0

5 - 5 - 0.97F4 = 0

- 0.97 F4 = 0

F4 = 0

Y - component

Sum of forces = F2 + F4 sin 15

Sum of forces = 0

5 + 0.26F4 = 0

0.26 F4 = -5

F4 = -5/0.26

F4 = -19.23 N

Simulating F4 back into the equation

Sum of forces = F1 - F3 - F4cos 15

- F4cos Ø = 0

- (-19.23) cos Ø = 0

Cos Ø = 0

Ø = 1

Does the angle formed approximate 15 degrees ? NO

7 0

3 years ago

Write the algebraic equation for the following: The sum of 4 and another number is 22.

s344n2d4d5 [400]

Answer:

$4+x=22$

Step-by-step explanation:

Hope this helps

5 0

1 year ago

Read 2 more answers

Someone please help me with #28 and #30 I got myself super confused.

bogdanovich [222]

<u><em>Question 28:</em></u>

$S-2\pi r^2=2\pi rh$ Subtract $2\pi r^2$ on both sides

$\frac{S-2\pi r^2}{2\pi r} =h$ Divide $2\pi r$ on both sides

<u><em>*(Answer)*=</em></u> $\frac{S-2\pi r^2}{2\pi r} =h$

<u><em>Question 30:</em></u>

$a*t=v_{1}+v_{0}$ Multiply $t$ on both sides

$a*t+v_{0} =v_{1}$ Subtract $v_{0}$ on both sides

<u><em>*(Answer)*=</em></u> $a*t+v_{0} =v_{1}$

Hope This Helps

BangtanBoyScouts

4 0

3 years ago

Express f(x) = |x-2| +|x+2| in the non-modulus form. Hence, sketch the graph of f.

alexgriva [62]

Recall that

$|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

There are three cases to consider:

(1) When $x+2$ , we have $|x+2|=-(x+2)$ and $|x-2|=-(x-2)$ , so

$|x-2|+|x+2|=-(x-2)-(x+2)=-2x-4$

(2) When $x+2\ge0$ and $x-2$ , we get $|x+2|=x+2$ and $|x-2|=-(x-2)$ , so

$|x-2|+|x+2|=-(x-2)+(x+2)=4$

(3) When $x-2\ge0$ , we have $|x+2|=x+2$ and $|x-2|=x-2$ , so

$|x-2|+|x+2|=(x-2)+(x+2)=2x$

So

$|x-2|+|x+2|=\begin{cases}-2x-4&\text{if }x$

4 0

2 years ago

Find the exact volume of the cylinder.

GaryK [48]

It is "(C) 225pi cm3"
I am 100% sure this is correct.

8 0

2 years ago