3 years ago

Ax^4+bx^2+c Factorize it

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

6 0

Answer:

ax^4 + bx^2 + c.

Step-by-step explanation:

I think you have to consider this as prime.

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Arlecino [84]

2 years and 6 months

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mark brainliest :)

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3 years ago

Read 2 more answers

a number, half of that number, and one-third of that number are added. The result is 22. What is the original number

aliina [53]

A number, half of that number, and one-third of that number are added.
This statement can be expressed as:
x + (1/2)x + (1/3)x

The total is 22
x + (1/2)x + (1/3)x = 22

So let's look for x
x + (1/2)x + (1/3)x = 22
(6x + 3x + 2x) / 6 = 22
11x = 132
x = 12

So the original number is 12.

7 0

3 years ago

Calculeaza a-b+c stiind ca a-1538=b+718=2000-c=1074

snow_tiger [21]

A-1538=1074
a=1538+1074
a=2612

b+718=1074
b=1074-718
b=356

2000-c=1074
c=2000-1074
c=926

2612-356+926=3182

8 0

3 years ago

How can help me with (c)

dimulka [17.4K]

X^2-y^3=6
x^2-6=y^3
y= (x^2-6)^1 by 3

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3 years ago

What is 5r^2 = 80 It says solve each equation with the quadratic formula.

katovenus [111]

$5r^2 = 80 \ \ /:5\\ \\r^2=\frac{80}{5}\\ \\r^2=16\\ \\r^2-16=0\\ \\ r^2 - 4^2 =0 \\ \\(r-4)(r+4)=0$

$r-4=0 \ \ \vee \ \ r+4 =0 \\ \\r=4 \ \ \vee \ \ r = -4 \\ \\ \\ a^2-b^2 = (a- b)(a+b)$

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3 years ago