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svet-max [94.6K]

3 years ago

6

Least to greatest 7.618,7.681,7.6801,7.0681

2 answers:

kramer3 years ago

4 0

Least to greatest 7.0681, 7.618, 7.6801,7.681

......................................................

maw [93]3 years ago

3 0

7.0681
7.618
7.681
7.6801

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What is the percentage 45cm of 2m

EleoNora [17]

22.5% is the correct answer. 45/200 = 225.

3 0

3 years ago

Bailey company purchased a new delivery truck for $35,000. the sales tax was $2,000. the logo of the company was painted on the

iragen [17]

To find the cost of the truck, you will need to add all the expenses that incurred in addition to the actual cost of the truck.

35000 + 2000 + 1200 + 120 + 220 = 38540

The cost including expenses should be recorded as $38540.

7 0

3 years ago

*What is the perimeter of a square with side length 13.4 inches? Write the direct variation equation.

Zarrin [17]

Answer:

53.6 inches

p = 4s

Step-by-step explanation:

The perimeter of a square is four times the side

The direct variation equation is

p = 4s

p = 4 * 13.4

p = 53.6 inches

5 0

3 years ago

Steve sells half of his comic books and then buys 6 more now he has 14 comic books

ddd [48]

Let b be the initial amount of books. Steve sells half of his books, so the number of Steve's books become

$b \to \dfrac{b}{2}$

Then, he buys 6 more, so the number increases by 6:

$\dfrac{b}{2} \to \dfrac{b}{2}+6$

This number is now 14, so you have

$\dfrac{b}{2}+6 = 14$

Subtract 6 from both sides:

$\dfrac{b}{2} = 8$

Multiply both sides by 2:

$b = 16$

4 0

3 years ago

Read 2 more answers

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

2 years ago