Ask question

Ask question

All categories

gayaneshka [121]

3 years ago

12

During a concert the temperature should not rise above 40°C.

1 answer:

zaharov [31]3 years ago

6 0

Answer:

the answer is 55

Step-by-step explanation:

if you subtract 40 degrees from 95 degrees you get 55 degrees. There are plenty of other ways to do that problem but i have a little bit of time. SORRY!

- I hope that helped

You might be interested in

Which net represents the pyramid with the greatest surface area

vladimir1956 [14]

Answer:

the top one that's orange

4 0

3 years ago

Need help with math problem if do get 5 star

11Alexandr11 [23.1K]

<h3><u>Answer</u><u>:</u></h3>

The point ( 22 , 23 ) lies in Ist quadrant

<h3><u>Explanation</u><u>:</u></h3>

The intersection of x and y axis divides the coordinate plane into 4 sections. These four sections are called quarrants. These quadrants are named as Roman numerals I, II, III and IV quadrant. The start with the top right corner and move in anti clockwise direction .

In a x y plane , both the values of x and y are positive in Ist quadrant

3 0

2 years ago

Find the quotient of 9 divided by 5/8 .

MissTica

9 / 5/8 =

9/1 / 5/8 =

9/1 * 8/5 = 72/5 = 14 2/5

5 0

3 years ago

Odd numbers 3 and 19

Lana71 [14]

Yes those are odd numbers

5 0

3 years ago

Read 2 more answers

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

Paha777 [63]

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,

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

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

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

We know that,

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

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

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

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

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

Using the trigonometric ratios, we get

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

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

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

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

6 0

3 years ago