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ruslelena [56]
2 years ago
13

Solve the system of equations. y=x+2 4x-4y= 1

Mathematics
2 answers:
Tpy6a [65]2 years ago
5 0

Answer:

No solution

Step-by-step explanation:

  1. 4x - 4y = 1
  2. 4x - 4(x + 2) = 1
  3. 4x - 4x - 8 = 1
  4. - 8 = 1

Our end result is -8 = 1. This is not true, so this system of equations have no solution.

Varvara68 [4.7K]2 years ago
5 0

Answer:

No solutions

Step-by-step explanation:

Step 1: Substitute x + 2 for y in 4x - 4y = 1.

  • 4x-4(x+2)=1
  • 4x+(-4)(x)+(-4)(2)=1
  • 4x-4x-8=1
  • -8 = 1  (simplify both sides)
  • -8+8=1+8
  • 0\neq 9

Thus, the answer is no solutions.

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(03.05 MC)
vagabundo [1.1K]

Answer:

I can not answer for some reason

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Step-by-step explanation:

6 0
2 years ago
Helppp plzzz solve them
Airida [17]
2. y=12000(.06)4x<span>
3. y=300(.08)5x
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5 0
3 years ago
Explain how to perform a​ two-sample z-test for the difference between two population means using independent samples with sigma
Rom4ik [11]

Answer:

Step-by-step explanation:

In the two independent samples application, it involves the test of hypothesis that is the difference in population means, μ1 - μ2. The null hypothesis is always that there is no difference between groups with respect to means.

Null hypothesis: ∪₁ = ∪₂. where ∪₁ represent the mean of sample 1 and ∪₂ represent the mean of sample 2.

A researcher can hypothesize that the first mean is larger than the second (H1: μ1 > μ2 ), that the first mean is smaller than the second (H1: μ1 < μ2 ), or that the means are different (H1: μ1 ≠ μ2 ). These ae the alternative hypothesis.

Thus for the z test:

    if n₁ > 30 and n₂ > 30

     z = X₁ - X₂ / {Sp[√(1/n₁ + 1/n₂)]}

where Sp is √{ [(n₁-1)s₁² + (n₂-1)s₂²] / (n₁+n₂-2)}

5 0
3 years ago
I need help for this, thanks
gulaghasi [49]

(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.

---------------------------------------------------------------------------------------

(b) The following functions are positive in quadrant III:

tangent, cotangent

The following functions are negative in quadrant III

cosine, sine, secant, cosecant

A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.

5 0
3 years ago
Can someone help me with this problem and give the answer
kykrilka [37]

Answer:

D

Step-by-step explanation:

<h2><u>No more means the other number is less than or equal to.</u></h2>
4 0
3 years ago
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