A toy store marks down every toy by 15% in January. How much does a toy cost during January? Use p for the price in December.

Mathematics

1 answer:

Vinil7 [7]2 years ago

6 0

Answer:

s=p−(p⋅d) . 0.75p

Step-by-step explanation:

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What is the simplified expression for 4 to the power of negative 3 multiplied by 3 to the power of 4 multiplied by 4 to the powe

blondinia [14]

Writing out: 4 to the power of negative 3 multiplied by 3 to the power of 4 multiplied by 4 to the power of 2 whole over 3 to the power of 5 multiplied by 4 to the power of negative 2

(4^(-3) * 3^4 * 4*2) / (3^5 * 4^(-2))

= (4^(-3+2) * 3^4) * 3^(-5) * 4^2

= 4^(1) * 3^(-1)

= 4/3

b. 4 over 3

8 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]

<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)

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

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