All categories

3 years ago

Please help I’ll give points

Mathematics

1 answer:

Marat540 [252]3 years ago

8 0

Answer:

$m = 360 - (59 + 77)$

You might be interested in

Help me asap i need help i dont understand

QveST [7]

Answer:

∠A + ∠B = 90°

Step-by-step explanation:

Complementary angles are angles which, combined, add to 90°. Therefore, an equation displaying the relationship between two complementary angles may look like this: ∠A + ∠B = 90°.

5 0

2 years ago

Cristina uses a ruler to measure the length of her math textbook. She says that the book is 4/10 meters long. Is her measurement

a_sh-v [17]

Answer:

2/5

Step-by-step explanation:

You could still divide by 2

4 0

3 years ago

Find area of triangle whose base is 2 √3 and height is 3 √3

Lera25 [3.4K]

Area of a triangle = base times height
A = 2√3 times 3√3
A = 18
The answer is 18
Hope that helps

7 0

3 years ago

John's commute time to work during the week follows the normal probability distribution with a mean time of 26.7 minutes and a s

STatiana [176]

Answer:

0.32493

Step-by-step explanation:

We have been given that John's commute time to work during the week follows the normal probability distribution with a mean time of 26.7 minutes and a standard deviation of 5.1 minutes.

We are asked to find the probability that the commute time for a randomly selected day will be between 28 and 34 minutes.

First of all, we will find z-score corresponding to 28 minutes and 34 minutes. Then, we will find area under normal curve between both z-scores.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{28-26.7}{5.1}=\frac{1.3}{5.1}=0.25$

$z=\frac{34-26.7}{5.1}=\frac{7.3}{5.1}=1.43$

Now, we need to find the probability between z-score of 1.43 and 0.25.

$P(0.25$ .

Using formula $P(a$ , we will get:

$P(0.25$

Therefore, the probability, that the commute time for a randomly selected day will be between 28 and 34 minutes, is 0.32493.

6 0

3 years ago

lim x rightarrow 0 1 - cos ( x2 ) / 1 - cosx The limit has to be evaluated without using l'Hospital'sRule.

zaharov [31]

Answer with Step-by-step explanation:

Given

$f(x)=\frac{1-cos(2x)}{1-cos(x)}\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-(cos^2{x}-sin^2{x})}{1-cos(x)})\\\\(\because cos(2x)=cos^2x-sin^2x)\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-cos^2x}{1-cos(x)}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}(\frac{(1-cosx)(1+cosx)}{1-cosx}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}((1+cosx)+\frac{sin^2x}{1-cosx})\\\\\therefore \lim_{x \rightarrow 0}f(x)=1$

6 0

3 years ago