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3 years ago

How many weeks is 18 days?

Mathematics

2 answers:

MrRissso [65]3 years ago

8 0

Answer:

Step-by-step explanation:

18/7=2.57

vaieri [72.5K]3 years ago

4 0

Answer:

ok so we know there are 7 days per week.

so now we are looking to see how many times 7 days (a week) can fit in 18

Step-by-step explanation:

18/7=2.6

so 2 weeks,

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27 is 72% of what number?

NNADVOKAT [17]

Answer:

37.5

Step-by-step explanation:

37.5 · 0.72 = 27

5 0

2 years ago

Read 2 more answers

In is shopping for a couch with her dad and hears him ask the salesperson, "How much is your commission?" The salesperson says t

Y_Kistochka [10]

Answer:

A. $27.23

B. $467.77

Step-by-step explanation:

Lin is shopping for a couch with her dad and heard him ask the salesperson “how much is your commission?” The salesperson says that her commission is 5 1/2% of the selling price. A.How much commission will the salesperson earn by selling a couch for $495?

B.How much money will the store get from the sale of the couch?

Salesperson commission = 5 1/2%

Price of couch = $495

A. Amount earned by the salesperson = 5 1/2% of $495

= 5.5% × 495

= 0.055 × 495

= 27.225

Approximately

Amount earned by the salesperson = $27.23

B.How much money will the store get from the sale of the couch?

Amount the store get = Total price - amount earned by salesperson

= $495 - $27.23

= $467.77

8 0

3 years ago

PLEASE HELP!!!!!!

vlada-n [284]

Answer:

I believe D. B12

5 0

3 years ago

Read 2 more answers

Write the equation of a polynomial of degree 3, with zeros 1, 2 and -1 where f(0)=2

drek231 [11]

<u>Answer:</u>

The equation of a polynomial of degree 3, with zeros 1, 2 and -1 is $x^{3}-2 x^{2}-x+2=0$

<u>Solution:</u>

Given, the polynomial has degree 3 and roots as 1, 2, and -1. And f(0) = 2.

We have to find the equation of the above polynomial.

We know that, general equation of 3rd degree polynomial is

$F(x)=x^{3}-(a+b+c) x^{2}+(a b+b c+a c) x-a b c=0$

where a, b, c are roots of the polynomial.

Here in our problem, a = 1, b = 2, c = -1.

Substitute the above values in f(x)

$F(x)=x^{3}-(1+2+(-1)) x^{2}+(1 \times 2+2(-1)+1(-1)) x-1 \times 2 \times(-1)=0$

$\begin{array}{l}{\rightarrow x^{3}-(3-1) x^{2}+(2-2-1) x-(-2)=0} \\ {\rightarrow x^{3}-(2) x^{2}+(-1) x-(-2)=0} \\ {\rightarrow x^{3}-2 x^{2}-x+2=0}\end{array}$