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valina [46]
3 years ago
13

Y=x-4 4x+y=26 Solve with substitution

Mathematics
1 answer:
Brums [2.3K]3 years ago
7 0

Answer:

(6,2)

Step-by-step explanation:

Since y=x-4, we can substitute this into the second equation so that there is only one variable that we can solve for.

4x+(x-4)=26

5x-4=26

5x=30

x=6

Now we can plug this back into the first equation.

y=(6-4)

y=2

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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
I'm having trouble on ixl here is what I have to do
Sophie [7]

Answer:

6(9r-7)

Step-by-step explanation:

54r–42

Both 54 and 43 can be divided by 6

54/6 =9  and 42/6 =7

6*9r - 6*7

Factor out the 6

6(9r-7)

8 0
3 years ago
Read 2 more answers
Simplify the following expression: 7y+2x-10y+3x^2-10x
vichka [17]
-3y-8x+3x^2 is the simplified version of 7y+2x-10y+3x^2-10x
7 0
3 years ago
A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 439.0 gram setting.
Vlada [557]
<h2>Answer with explanation:</h2>

Let \mu be the average weight of chocolate chips in each bag.

As per given , we have

H_0: \mu =439.0\\\\ H_a: \mu

Since H_a is left-tailed , so we perform a left-tailed test.

Also, the population standard deviation is known \sigm=21.0.

So we use z-test.

Test statistic : z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqt{n}}}

Substitute given value ,\overline{x}=433.0\ , n=47,\ \sigma=21 \ , \mu=439

z=\dfrac{433-439}{\dfrac{21}{\sqrt{47}}}\approx-1.96

Significance level = 0.05

Decision rule : Reject H_0 when p-value < 0.05.

By z-table , P- value for left-tailed test  = P(z<-1.96)= 1- P(z<1.96)

[∵ P(Z<-z)=1-P(Z<z)]

= 1- 0.9750

=0.025

Since , P-value(0.025) < 0.05 , so we reject the null hypothesis.

We support the claim that the machine is underfilling the bags at 0.05 significance level .

3 0
3 years ago
Angie and Becky each completed a separate proof to show that the measures of vertical angles AKG and HKB are equal. Who complete
BigorU [14]

Linear pair angles, which are angles that together form a straight line are

supplementary angles.

The one that completed the proof incorrectly is <u>Becky</u>.

Reasons:

The two column proof is presented as follows;

Statement {}                                                             Reason

1. Segment GH intersects segment AB at K {}     1. Given

2. m∠AKG + m∠HKB = 180°    {}            2.Definition of Supplementary Angles

m∠GKB + m∠HKB = 180°

3. m∠AKG + m∠HKB = m∠GKB + m∠HKB      {}  3. Substitution property

4. m∠AKG = m∠HKB     {}                                      4. Subtraction Property

The difference between Angie's Proof and Becky's Proof is in Statement 2.

  • Angie states that; m∠AKG + m∠HKB = 180° and m∠GKB + m∠HKB = 180° by definition of Supplementary Angles

  • Becky states that; m∠AKG + m∠HKB = 180° and m∠GKB + m∠HKB = 180° by Angle Addition Postulate

Becky's proof is incorrect because the measure of angles  m∠AKG and

m∠HKB and m∠GKB and m∠HKB are not given, therefore, the use of the

reason of Angle Addition Postulate in statement 2. is incorrect.

Learn more here:

brainly.com/question/13204208

8 0
3 years ago
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