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3 years ago

10 × 980 what's the awnser

Mathematics

2 answers:

egoroff_w [7]3 years ago

8 0

Add a zero to the end I the value and you would get 9800

weeeeeb [17]3 years ago

4 0

9800 hope this helps idk what else to type

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Clint's salary increased from $24000 to $36000 over a three-year period. Helen's salary increased from $29000 to $43, 500 ov

lesya [120]

Answer:

Helen's salary increased more.

Step-by-step explanation:

We know that Clint's salary increased from $24000 to $36000 over a three-year period.

So the increase in Clint's salary will be = 36000 - 24000 = $12000

While Helen's salary increased from $29000 to $43, 500 over the same period of 3 years.

So the increase in Helen's salary will be = 43500 - 29000 = $14500

Therefore, in absolute terms, Helen's salary has increased more.

8 0

4 years ago

What is the simplified form of the following expression? 6c^2 + 2.5d - d + 2c^2 - 3d

Airida [17]

Simplify the Expression.

8c^2 - 1.5d

4 0

4 years ago

Lola and Brandon had a running race. They ran 3/10 of a mile. Lola was in the lead for 4/5 of the distance. For what fraction of

Kruka [31]

Answer:

Lola was in lead for $\frac{6}{25}$ of a mile.

Step-by-step explanation:

We have been given that Lola and Brandon had a running race. They ran 3/10 of a mile. Lola was in the lead for 4/5 of the distance.

To find the fraction of a mile for which Lola was in the lead, we need to find 4/5 of 3/10 as:

$\text{Fraction for which Lola was in lead}=\frac{3}{10}\times\frac{4}{5}$

$\text{Fraction for which Lola was in lead}=\frac{3}{5}\times\frac{2}{5}$

$\text{Fraction for which Lola was in lead}=\frac{3\times2}{5\times5}$

$\text{Fraction for which Lola was in lead}=\frac{6}{25}$

Therefore, Lola was in lead for $\frac{6}{25}$ of a mile.

7 0

3 years ago

cynthia invests some money in a bank which pays 5% compound interest per year. She wants it to be worth over £8000 at the end of

exis [7]

Answer:

The smallest amount is $\£6,911$

Step-by-step explanation:

we know that

The compound interest formula is equal to

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

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=3\ years\\ A=\£8,000\\ r=5\%=5/100=0.05\\n=1$

substitute in the formula above

$8,000=P(1+\frac{0.05}{1})^{1*3}$

solve for P

$8,000=P(1.05)^{3}$

$P=8,000/(1.05)^{3}$

$P=\£6,910.70$

Round to the nearest pound she can invest

The smallest amount is $\£6,911$

6 0

3 years ago

1. Find the least common multiple. 20 and 12

musickatia [10]

Answer:

answer 60

Step-by-step explanation:

The LCM(12, 20) = 60.

4 0

3 years ago