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Thepotemich [5.8K]
3 years ago
6

Plz

Mathematics
1 answer:
allsm [11]3 years ago
4 0
Top box is 2x
bottom box is 3
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(x-2)(-5x^2+x)=(x)(-5x^2)+(x)(x)+(-2)(-5x^2)+(-2)(x) is an exsample of
e-lub [12.9K]

this is an example of linear equations is one variable

6 0
3 years ago
PLS GIVE THE ANSWER!!!!!! ALSO ITS NOT A NUMBER IS IT C OR A
Natasha2012 [34]

Answer:

Where is the picture or questions?

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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
2 years ago
Someone please help me! Easy 10 points
Katyanochek1 [597]

Answer:

I can't see ;-; show close up

5 0
2 years ago
What are 3 completed examples for functions?
Lyrx [107]
Example 1

Write y = x2 + 4x + 1 using function notation and evaluate the function at x = 3.

Solution

Given, y = x2 + 4x + 1

By applying function notation, we get

f(x) = x2 + 4x + 1

Evaluation:


Substitute x with 3

f (3) = 32 + 4 × 3 + 1 = 9 + 12 + 1 = 22

Example 2

Evaluate the function f(x) = 3(2x+1) when x = 4.

Solution

Plug x = 4 in the function f(x).

f (4) = 3[2(4) + 1]

f (4) = 3[8 + 1]

f (4) = 3 x 9

f (4) = 27

Example 3

Write the function y = 2x2 + 4x – 3 in function notation and find f (2a + 3).

Solution

y = 2x2 + 4x – 3 ⟹ f (x) = 2x2 + 4x – 3


Substitute x with (2a + 3).

f (2a + 3) = 2(2a + 3)2 + 4(2a + 3) – 3

= 2(4a2 + 12a + 9) + 8a + 12 – 3
= 8a2 + 24a + 18 + 8a + 12 – 3
= 8a2 + 32a + 27
3 0
2 years ago
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