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3 years ago

Y= 2x2 – 9x + 8 Deter sin the number of solutions for quadratic equation

Mathematics

1 answer:

artcher [175]3 years ago

7 0

Well if it is a quadratic equation, then it has 2 solutions

Do you need the solutions or just the number of solutions?

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A cell phone costs 120 dollars. The price reduced 30% for a sale. How much is the savings

Licemer1 [7]

Answer:

You save 36 dollars.

Step-by-step explanation:

30% of 120 is 36.

You spend $84

5 0

3 years ago

How to figure a composite figure area

photoshop1234 [79]

Answer:

To find the area of a composite figure or other irregular-shaped figure, divide it into simple, nonoverlapping figures. Find the area of each simpler figure, and then add the areas together to find the total area of the composite figure.

Step-by-step explanation:

Did this help, do you need an example

8 0

3 years ago

What's the solution to these questions?

sveticcg [70]

1st Question:
X = 18

2nd Question:
X=20

6 0

3 years ago

What are the zeros of the polynomial function? f(x)=x^3−x^2−4x+4

harkovskaia [24]

Answer: (B) -2, (E) 1, (F) 2

<u>Step-by-step explanation:</u>

x³ - x² - 4x + 4

= x²(x - 1) - 4(x - 1)

= (x² - 4) (x - 1)

= (x - 2)(x + 2)(x - 1)

Set each factor equal to zero to find the roots:

x - 2 = 0 x + 2 = 0 x - 1 = 0

x = 2 x = -2 x = 1

5 0

3 years ago

Using the method of mathematical induction to prove that equalities are true for values of n indicated:

Dima020 [189]

$2^2+4^2+6^2+...(2n)^2=\frac{2n(n+1)(2n+1)}{3};\ n\geq1\\\\chek\ for\ n=1:\\L=2^2=4;\ R=\frac{2\cdot1(1+1)(2\cdot1+1)}{3}=\frac{2\cdot2\cdot3}{3}=4\\L=R\\-----------------------\\
assumption\ for\ n=k\\2^2+4^2+6^2+...+(2k)^2=\frac{2k(k+1)(2k+1)}{3}\\-----------------------\\thesis\ for\ n=k+1\\2^2+4^2+6^2+...+(2k)^2+[2(k+1)]^2=\frac{2(k+1)(k+1+1)[2(k+1)+1]}{3}\\-----------------------$
$proff:\\L=2^2+4^2+6^2+...+(2k)^2+(2k+2)^2=\frac{2k(k+1)(2k+1)}{3}+(2k+2)^2\\\\=\frac{(2k^2+2k)(2k+1)}{3}+\frac{3(2k+2)^2}{3}=\frac{4k^3+2k^2+4k^2+2k+3(4k^2+8k+4)}{3}\\\\=\frac{4k^3+6k^2+2k+12k^2+24k+12}{3}=\boxed{\frac{4k^3+18k^2+26k+12}{3}}\\\\R=\frac{2(k+1)(k+1+1)[2(k+1)+1]}{3}=\frac{(2k+2)(k+2)(2k+2+1)}{3}\\\\=\frac{(2k^2+4k+2k+4)(2k+3)}{3}=\frac{(2k^2+6k+4)(2k+3)}{3}=\frac{4k^3+6k^2+12k^2+18k+8k+12}{3}\\\\=\boxed{\frac{4k^3+18k^2+26k+12}{3}}\\\\L=R$

4 0

3 years ago