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2 years ago

What is negative 3/4 minus 1/8

Mathematics

1 answer:

KATRIN_1 [288]2 years ago

4 0

Answer:

-7/8

Step-by-step explanation:

To get the answer you have to convery the factor to the same denominator. Which is 8. To it would be -6/8-1/8 which give you -7/8

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Evaluate the expression:<br><br> 2/3-1/4+(-5/6)

Aliun [14]

Answer:

$\\ \frac{2}{3} - \frac{1}{4} + \frac{( - 5)}{6}$

$\\ = \frac{2}{3} - \frac{1}{4} - \frac{5}{6}$

$\\ = \frac{8 - 3 - 10}{12}$

$\\ = \frac{8 - 13}{12}$

$\\ = \frac{ - 5}{12}$

3 0

2 years ago

A fraction is written in simplest form when the GCF of the<br> numerator and the denominator is

Natasha_Volkova [10]

Answer:

simplified

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

At the start of 2014, Mike's car is worth 12000. the car depreciates by 30 percent every year. how much is his car worth in 2017

Anarel [89]

Answer:

$4,116

Step-by-step explanation:

Worth of Mike's car at the start of 2014 = $12,000

If the car is said to depreciates every year by 30% = 30/100 = 0.3

The worth of the car at the start of 2017 is what we are to determine.

This means that the car depreciated by 30% (0.3) for 3 years since 2014 (2017 - 2014 = 3 yrs)

The worth at the start of 2017 would be calculated as follows:

12,000 × (1 - 0.3)³

= 12,000 × (0.7)³

= 12,000 × 0.343

= 4,116

Worth of the car at the start of 2017 would be $4,116

4 0

3 years ago

Which is bigger 100 or 10%

Elanso [62]

Answer: 10%

Step-by-step explanation:

even tho 100 is a bigger number the amount of 10% is 10 times bigger than 100%

4 0

2 years ago

Read 2 more answers

A fund earns a nominal rate of interest of 6\% compounded every two years. Calculate the amount that must be contributed now to

Nesterboy [21]

Answer:

$712.

Step-by-step explanation:

We have been given that a fund earns a nominal rate of interest of 6% compounded every two years. We are asked to find the amount that must be contributed now to have 1000 at the end of six years.

We will use compound interest formula to solve our given problem.

$A=P(1+\frac{r}{n})^{nt}$ , where,

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

$6\%=\frac{6}{100}=0.06$

Since interest is compounded each two years, so number of compounding per year would be 1/2 or 0.5.

$1000=P(1+\frac{0.06}{0.5})^{0.5*6}$

$1000=P(1+0.12)^{3}$

$1000=P(1.12)^{3}$

$1000=P*1.404928$

$\frac{1000}{1.404928}=\frac{P*1.404928}{1.404928}$

$P=711.7802478$

$P\approx 712$

Therefore, an amount of $712 must be contributed now to have 1000 at the end of six years.

7 0

3 years ago