$\dfrac{3}{4}+\dfrac{7}{8}=\dfrac{3\cdot2}{4\cdot2}+\dfrac{7}{8}=\dfrac{6}{8}+\dfrac{7}{8}=\dfrac{6+7}{8}=\dfrac{13}{8}=\dfrac{8+5}{8}=\dfrac{8}{8}+\dfrac{5}{8}=1\dfrac{5}{8}\\\\Answer:\boxed{1\frac{5}{8}\ hour.}$

5 0

3 years ago

The sum of two and four and six times eight?

MissTica

Answer:

2+4 6(8)

Step-by-step explanation:

pls give brainliest

4 0

3 years ago

Read 2 more answers

If you remove the last digit (one’s place) from a 4-digit whole number, the resulting number is a factor of the 4-digit number.

shutvik [7]

Answer:

900

Step-by-step explanation:

We assume that your 4-digit number must be in the range 1000 to 9999. Clearly, any number ending in zero will meet your requirement:

1000/100 = 10

3890/389 = 10

However, the requirement cannot be met when the 1s digit is other than zero.

For some 3-digit number N and some 1s digit x, the 4-digit number will be

4-digit number: 10N+x

Dividing this by N will give ...

(10N+x)/N = 10 remainder x

N will only be a factor of 10N+x when x=0.

So, there are 900 4-digit numbers that meet your requirement. They range from 1000 to 9990.

6 0

3 years ago