All categories

3 years ago

What is one quarter of 5 hours

Mathematics

2 answers:

Ivahew [28]3 years ago

6 0

Answer:

75 minutes

Step-by-step explanation:

5 hours = 60 x 5 = 300 minutes

300 ÷ 4 = 75 minutes

vitfil [10]3 years ago

3 0

Answer:

15 minutes

There are 20 quarter hours in 5 hours. This is solved by first breaking down an hour into quarters. A quarter is 1/4 of an hour, or 15 minutes.

You might be interested in

On a certain day, the company took 88 people on whale watching trips. There were 44 children aged twelve and under, of which som

Setler [38]

"(44 – x)35 + (88 – 44) ∙ 46 = C" is the one equation among the following choices given in the question that can be used to <span>find the value of x, where C represents the total amount of money collected from tickets that day. The correct option among all the options that are given in the question is the third option.</span>

6 0

2 years ago

Read 2 more answers

Can I take a picture of the problem and upload for you to see?

eduard

Yes you can …………………………

6 0

2 years ago

Can anyone help me?pls

Mama L [17]

Answer/Step-by-step explanation:

Given:

C = right angle = 90°

BC = a = ?

AB = c = 12

AC = b = 9

Required:

a, <A, and <B

Solution:

✔️Find a using Pythagorean Theorem:

a = √(c² - b²)

a = √12² - 9²) = √63

a = 7.93725393 = 8 (nearest whole number)

✔️Find A by applying trigonometric ratio:

Thus,

Reference angle = A

Hypotenuse = 12

Adjacent = 9

Therefore,

$cos(A) = \frac{adj}{hyp} = \frac{9}{12}$

$cos(A) = 0.75$

$A = cos^{-1}(0.75)$

m<A = 41° (nearest whole number)

✔️Find B:

m<B = 180 - (90 + 41) (sum of interior angles of triangle)

m<B = 49°

8 0

3 years ago

Which is the solution of LaTeX: \frac{t}{2.5}=-5.2

Inga [223]

Answer:

this is too hard which class are u in

3 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