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2 years ago

Can someone please check if this is correct?

Mathematics

1 answer:

Solnce55 [7]2 years ago

7 0

Answer:

Just two corrections! See attached image.

Step-by-step explanation:

The product of 8 and b taken from 10 means begin with 10 and subtract 8b from it.

8 times the difference of 10 and b means find 10 - b first, then multiply by 8.

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75/12=□/4=160/□what is the answer?

Vlada [557]

Answer:

75/12= 6.25/4=160/102.4

4 0

3 years ago

What is the value of the expression when n = 3? GO - 2 n(5 + n-8-3n)

lukranit [14]

Answer:

Step-by-step explanation:

since n = 3

= -2n ( 5n+n-8-3n)

= -2 (3) ( 5+(3)-8-3(3) )

= -6 (5+3-8-9)

= -6 (5-5-9)

= -6 (-9)

= 54.

8 0

3 years ago

Suppose c was a total membership cost for m the two equations that model c are c=30m+50 an c=20m+100

Maksim231197 [3]

Answer:

5 months

Step-by-step explanation:

Given

$c = 30m + 50$

$c = 20m + 100$

Required [Missing from the question]

At what month will the cost be equal

To do this, we equate both expressions

$c = c$

This gives:

$30m + 50 = 20m + 100$

Collect like terms

$30m - 20m = 100 - 50$

$10m = 50\\$

Solve for m

$m = \frac{50}{10}$

$m = 5$

8 0

3 years ago

Two solid metal spheres have masses of 5 kg and 135 kg respectively. If the radius of the smaller one is 4 cm, find the radius o

Cerrena [4.2K]

Answer: 12 cm

Step-by-step explanation:

The masses of the spheres are proportional to their volums, and the cube of the ratio is k^3=135/5=27 ==> k=3

The greater radius is 3*4=12 (cm)

8 0

2 years ago

At the beginning of each of her four years in college, Miranda took out a new Stafford loan. Each loan had a principal of $5,500

kaheart [24]

Answer:

D. $31,337.27

Step-by-step explanation:

We have that the initial amount of the loan is $5500.

Miranda took the loan for 4 years. So, the total present value is $5500×4 = $22,000.

The rate of interest on the loan is 7.5% i.e. 0.075 and it was for the duration of 10 years.

Also, it is given that the loan was compounded annually.

We have the formula as,

$P=\frac{\frac{r}{n}\times PV}{1-(1+\frac{r}{n})^{-t\times n}}$

i.e. $PV=\frac{P\times [1-(1+\frac{r}{n})^{-t\times n}]}{\frac{r}{n}}$

Substituting the values, we get,

i.e. $PV=\frac{P\times [1-(1+\frac{0.075}{12})^{-10\times 12}]}{\frac{0.075}{12}}$

i.e. $22000=\frac{P\times [1-(1+0.00625)^{-120}]}{0.00625}$

i.e. $22000=\frac{P\times [1-(1.00625)^{-120}]}{0.00625}$

i.e. $22000=\frac{P\times [1-0.4735]}{0.00625}$

i.e. $22000=\frac{P\times 0.5265}{0.00625}$

i.e. $P=\frac{22000\times 0.00625}{0.5265}$

i.e. $P=\frac{137.5}{0.5265}$

i.e. $P=261.16$

Thus, the total lifetime cost to pay of the loans compounded annually = 261.16 × 120 = $31,339.2

Hence, the total cost close to the answer is $31,337.27

7 0

3 years ago

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