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3 years ago

Lisa bought a suit on sale for 459. This price was 15% less than the original price. What was the original price?

1 answer:

7 0

If 459 is 15% less than the original, this means the original was 15% more than 459.

So we have to find 15% more of 459, and to do this, we have to:

Add 100 + 15 = 115
115/100 = 1.15
1.15 x 459 = 527.85

Hope this answers your question

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Need hep with this question

choli [55]

Answer:

V= -2000t + 20000

Step-by-step explanation:

Let m be the slope of our function :

m=(20000-16000)/(0-2)= -2000
so our function is V= -2000t + b
b is the intitial price so b = 20000
then : V= -2000t + 20000

Let's check :

take the second year : V= -2000*2 +20000 = 16000
that's true !

6 0

3 years ago

Helppppp needed plz quick

olga nikolaevna [1]

Answer:

Step-by-step explanation:

9x - 77 = 7x - 43

2x - 77 = -43

2x = 34

x = 17

9(17) - 77 = 153 - 77 = 76

7(17) - 43 = 119 - 43 = 76

z + 76 = 180

z= 104

4 0

3 years ago

-. Find the value of x. Round to the nearest degree.

Mice21 [21]

X is 44.4 degrees and round it off to the nearest degree is 44

7 0

2 years ago

Are the graphs of the lines in the pair parallel? Explain. Find the slope of a line parallel to 5x+2y=6

Sonja [21]

Answer:

Yes the graphs of the lines are in pair parallel.

Step-by-step explanation:

They are in pair parallel because they do not touch each other. To find the slope of the line that is parallel to 5x+2y=6...

The first thing to do is subtract 5x since its a positive so it would look like this:

5x+2y=6

-5x =-5x

Next you would divide by 2 on each side so the y can be isolated:

5x+2y=6

-5x =-5x

2y = -5x + 6

2 2

Lastly the 2 divided by 2 gets crossed out because they cancel each other out and divide -5x by 2 and 6 by 2 then you would be stuck with this:

y= -2.5x+3

So your parallel slope would be -2.5x.

8 0

3 years ago

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

A.

The student made an error in step 3 because a is positive in Quadrant IV; therefore,

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

The student's mistake is that a is positive in quadrant iv and his error is in step 3

3 0

3 years ago