All categories

2 years ago

Please help!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Valentin [98]2 years ago

7 0

Answer:

I am not sure but think that the answer is 1/20

victus00 [196]2 years ago

3 0

I believe you convert the whole number into a mixed fraction.

5 converted into quarters is 20/4. Then you divided 1/4 by 20/4.

1/4 divided by 20/4 is 1/20 or 0.05.

Please, don't refrain to tell me if this is incorrect. Thank you.

You might be interested in

One positive integer is 5 times another positive integer and their product is 320. What are the positive integers?

NeTakaya

Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
x = 5y
xy = 320

Substitute the first equation into the second equation
(5y)(y) = 320
5y^2 = 320
y^2 = 64
y = 8 (y must be positive)

The integers are (x, y) = (40, 8).

7 0

2 years ago

What is the most logical first step in solving the equation x2+6x+9=10?

valentina_108 [34]

You need to make the right side = 0
To do this you subtract 10 from both sides of the equation:-

x^2 + 6x + 9 - 10 = 10 - 10
x^2 + 6x - 1 = 0

5 0

3 years ago

Tommy is 22 years older than his son, Patrick. If their ages add up to 50, how old is Tommy?

alexandr1967 [171]

50 - 22 = 28
Tommy is 28

3 0

3 years ago

Avery bought a flute that cost $430. She paid a 6 1/2% sales tax rate. What was the total cost of the flute?

Georgia [21]

Answer:

457.95

Step-by-step explanation:

430 + 6.5% (27.95) = 457.95

3 0

3 years ago

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

drek231 [11]

Answer:

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

Solution:

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)

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

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

So, the equation is $x^{3}-2 x^{2}-x+2=0$

Let us put x = 0 in f(x) to check whether our answer is correct or not.

$\mathrm{F}(0) \rightarrow 0^{3}-2(0)^{2}-0+2=2$

Hence, the equation of the polynomial is $x^{3}-2 x^{2}-x+2=0$

3 0

3 years ago