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GaryK [48]
3 years ago
14

The price of an iPod dropped from $299.99 to $180.55. What was the percent decrease in price? (Round to nearest hundredth percen

t.)
Mathematics
2 answers:
ziro4ka [17]3 years ago
5 0
119.44  is that the whole question

anzhelika [568]3 years ago
3 0
The price dropped 299.99 - 180.55 = 119.44. So we need to find out what percentage 119.44 is of 299.99. 

<span>"119.44 is what percentage of 299.99" translates to </span>
<span>119.44 = x * 299.99 </span>

<span>So x = 119.44 / 299.99 = about 0.3981466, which is 39.81% when rounded to the nearest hundredth of a percent.</span>
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Draw an angle of negative pie over three in<br> Standard position.
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Check the picture below.

recall that negative angles go "clockwise".

7 0
3 years ago
A line has a slope of -6 over 7. What is the slope of the line parallel to it? And what is the slope of the line perpendicular t
olga_2 [115]
The slope of the parallel line is -6/7
the slope of the perpendicular line is 7/6
the slope of the line = -6/7
the gradient of two parallel lines are equal
the product of the gradient of two perpendicular lines is -1
: the gradient m1*m2 = -1
m2= -1(-6/7)
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3 0
3 years ago
9x + 26 + 7x - 17 = 2x + (-3x) + 5x<br><br> 16x + 9 = 0<br> 16x + 11 = 4x<br> 16x + 9 = 4x
zzz [600]

Answer:

C. 16x+9=4x

Step-by-step explanation:

4 0
3 years ago
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Four years ago, you started a job with an annual salary of $40,000. At the end of each year on the job, you received a raise of
Marrrta [24]

Answer:

the third one

Step-by-step explanation:

3 0
3 years ago
A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

               =\frac{24\times (-13)}{\sqrt{676}}

               =\frac{24\times (-13)}{26}

               = -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0
3 years ago
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