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antiseptic1488 [7]

2 years ago

11

QUESTION 1 of 10: The cost of a garment was determined to be $12. 50. You have been instructed to lower the cost to $10 per garm

ent. What
is percentage you were asked to reduce the cost by?

1 answer:

Zarrin [17]2 years ago

5 0

Answer:

20%

Step-by-step explanation:The answer is minus 2.50 so if you plug 12.50 times .2 you get your answer

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Please help i'm struggling with this!It's urgent!

lukranit [14]

Answer:

notation I believe

Step-by-step explanation:

6 0

3 years ago

A navigator plots the course a plane is currently traveling. The plane is 300 miles from its destination. If it continues on its

inna [77]

We have to use the rule of cosx° to solve this problem. Attached is a diagram of the navigator's course for the plane. It is similar to the shape of a triangle. We know the plane is 300 miles from its destination, so that will be one of the sides. On the current course, it is 325 miles from its destination, so that will be another one of the sides. The last side is 125 because that is the distance between the destination and the anticipated arrival. Cosx° is what we are looking for.

To find how many degrees off course the plane is, we must use the rules of Cosx°, which is shown in the attached image.

The plane is approximately 23° off course.

3 0

4 years ago

Find the average value fave of the function f on the given interval. f() = 5 sec2(/6), 0, 3 2

Inessa05 [86]

Answer:

$f_{ave}$ = ${\frac{10 }{3\pi }$

Step-by-step explanation:

To find - Find the average value fave of the function f on the given interval. f(x) = 5 sec²(x/6), [0, 3 $\pi$ /2]

Proof -

We know that,

Average value of f in the interval [a, b] is -

$f_{ave}$ = $\frac{1}{b - a}\int\limits^b_a {f(x)} \, dx$

Now,

Here a = 0, b = $\frac{3\pi }{2}$ , f(x) = $5sec^{2}(\frac{x}{6} )$

Now,

$f_{ave}$ = $\frac{1}{\frac{3\pi }{2} - 0}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx$

= ${\frac{2 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx$

= ${\frac{10 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {sec^2 ({\frac{x}{6} }) } \, dx$

= ${\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) }|^{\frac{3\pi }{2} }_0 \, ]$

= ${\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) - tan({0) } \, ]$

= ${\frac{10 }{3\pi }}[ {1 - 0 } ]$

= ${\frac{10 }{3\pi }$

⇒ $f_{ave}$ = ${\frac{10 }{3\pi }$

3 0

3 years ago

Estimate the percentage of never married American female ages from 25 to 29 in 2010

Andru [333]

in 1980 it was about 20%

in 1990 it was about 30%

about a 10% increase

in 2000 it was just under 40% so almost a 10% increase

so for estimating you can say it is increasing by about 10% every 10 years

so in 2010 you can estimate about 50% are unmarried

8 0

3 years ago

12.5-0.009=<br> 1.25-0.009=<br> 0.75-0.009=<br> 1.20-0.009=

tia_tia [17]

Hello! Here are your Answers:

12.5 - 0.009 = <u>12.491</u>

1.25 - 0.009 = <u>1.241</u>

0.75 - 0.009 = <u>0.741</u>

1.20 - 0.009 = <u>1.191</u>

<u>I hope This Helps!</u>

6 0

3 years ago