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Temka [501]
3 years ago
5

Directions: Simplify the following expressions. 2) 3 1) (40*2) - (6*11)

Mathematics
2 answers:
Over [174]3 years ago
8 0

Answer:

(40 x 2) - (6 x 11) (80 - 66) = 14

Aleksandr [31]3 years ago
5 0

Answer:

[40 × 2] - [6 × 11]

[80]-[66]

80-66

=14

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What is this answer I’m having a hard time
docker41 [41]

Answer:

1)x=4

2)

Step-by-step explanation:

12x-14=34

step1:     +14 on both sides

step2: 12x=48

          divide 12 from both sides

step3:x=4


6 0
3 years ago
Arthur’s employer withheld $15987.76 in federal income tax. After completing his return, Arthur has determined that his tax is $
Lerok [7]
I don’t know the answer
6 0
2 years ago
Needed help! Answer when possible.
Lera25 [3.4K]

Answer:

The word "solution" means an action or process of solving a problem

99 is a solution to 1/9x = 11 because  multiplying both side of the equation by 9 makes x = 99

3 0
3 years ago
Debra plans to invest $2,250 for 10 years. She can invest in a savings account that pays 4% simple intrest or a savings account
Diano4ka-milaya [45]

Answer:

\$180.55

Step-by-step explanation:

step 1

<u><em>Simple interest</em></u>

we know that

The simple interest formula is equal to

A=P(1+rt)

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

t=10\ years\\ P=\$2,250\\r=4\%=4/100=0.04

substitute in the formula above

A=2,250(1+0.04*10)

A=2,250(1.4)

A=\$3,150

step 2

<u><em>Interest compounded annually</em></u>

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=10\ years\\ P=\$2,250\\r=4\%=4/100=0.04\\n=1

substitute in the formula above

A=2,250(1+\frac{0.04}{1})^{1*10}  

A=2,250(1.04)^{10}  

A=\$3,330.55

step 3

Find the differences between the two final amounts

A=\$3,330.55-\$3,150=\$180.55

5 0
3 years ago
From 1960 to 2010, a certain money stock measure was growing at the rate of approximately 43e(1/2)x billion dollars per decade,
natulia [17]

Answer:

1222 billion dollars.

Step-by-step explanation:

To find the total increase from 1960 to 2010, we need to find the growth of each decade and sum them all:

In the period 1960-1970, we have x = 1, and the growth is:

y(1) = 43e(1/2) = 70.895

In the period 1970-1980, we have x = 2, and the growth is:

y(2) = 43e(2/2) = 116.8861

The growth in the following 3 periods are:

y(3) = 43e(3/2) =192.7126

y(4) = 43e(4/2) = 317.7294

y(5) = 43e(5/2) =523.8472

So the total growth in the period 1960 - 2010 is:

Total = y(1) + y(2) + y(3) + y(4) + y(5)

Total = 70.895 + 116.8861+192.726+317.7294+523.8472

Total = 1222.08\ billion\ dollars

Rounding to the nearest billion dollars, we have a total of 1222 billion dollars.

6 0
3 years ago
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