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Fantom [35]
3 years ago
10

Is -x^2+5x-21 linear or not?

Mathematics
2 answers:
Allushta [10]3 years ago
7 0
No, x is raised to the second power. i hope i helped!
MA_775_DIABLO [31]3 years ago
3 0
It is not linear because x is raised to the second power. Therefore it is not straight and would be kindof like a rainbow cut in half on the graph..
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3(2x-14) x=15-(-9x-5) solution?
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The radius of one sphere is twice as great as the radius of a second sphere. Find the ratio of their volumes. 1/2 1/4 1/8 1/16
Anton [14]

Let's pick some simple points with which to set up an example for ourselves for this. Let's let the smaller radius be 1, and the larger, twice that, be 2. The radius itself is a single unit measure; in other words, it's measured as inches, feet, cm, etc., while the volume is a cubed measure. Volume is measured in inches cubed, feet cubed, cm cubed, etc. Therefore, if we have the radii measuring 1:2, we simply cube those single unit measures to find the ratio of their volumes. 1 cubed is 1, and 2 cubed is 8. So your answer for this is 1/8.

5 0
4 years ago
Read 2 more answers
PLEASE HELP ASAP WILL GIVE BRAINLIEST !!!
My name is Ann [436]

Answer:

7/3

Step-by-step explanation:

7/3

8 0
3 years ago
What two rational expressions sum to 2x+3/x^2-5x+4
Anni [7]

Answer:

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}

Step-by-step explanation:

Given the rational expression: \frac{2x + 3}{x^2 - 5x + 4}, to express this in simplified form, we would need to apply the concept of partial fraction.

Step 1: factorise the denominator

x^2 - 5x + 4

x^2 - 4x - x + 4

(x^2 - 4x) - (x + 4)

x(x - 4) - 1(x - 4)

(x- 1)(x - 4)

Thus, we now have: \frac{2x + 3}{(x- 1)(x - 4)}

Step 2: Apply the concept of Partial Fraction

Let,

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}

Multiply both sides by (x - 1)(x - 4)

\frac{2x + 3}{(x- 1)(x - 4)} * (x - 1)(x - 4) = (\frac{A}{x- 1} + \frac{B}{x - 4}) * (x - 1)(x - 4)

2x + 3 = A(x - 4) + B(x - 1)

Step 3:

Substituting x = 4 in 2x + 3 = A(x - 4) + B(x - 1)

2(4) + 3 = A(4 - 4) + B(4 - 1)

8 + 3 = A(0) + B(3)

11 = 3B

\frac{11}{3} = B

B = \frac{11}{3}

Substituting x = 1 in 2x + 3 = A(x - 4) + B(x - 1)

2(1) + 3 = A(1 - 4) + B(1 - 1)

2 + 3 = A(-3) + B(0)

5 = -3A

\frac{5}{-3} = \frac{-3A}{-3}

A = -\frac{5}{3}

Step 4: Plug in the values of A and B into the original equation in step 2

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}

7 0
3 years ago
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