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aleksandrvk [35]
3 years ago
13

5x-2y+3z-2x-y+4z solve

Mathematics
2 answers:
Vaselesa [24]3 years ago
6 0
I'm not exactly sure how to approach this problem because it does not have an output, but we can combine like terms. 
Add all the x, y, and z together with the numbers.
3x-3y+7z is the final answer.
Jlenok [28]3 years ago
6 0
<span><span>3x-</span><span>3y</span></span>+<span>7<span>z, would be the answer because you have to group like terms so first off you would take 5x-2x which will give you 3x then you take -2y-y which gives you -3y so that you would have 3x-3y and then you would take 3z+4z which would give you 7z therefore the solution to the system is 3x-3y+7z

</span></span>
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Find the p​-value in a test of the claim that the mean College Algebra final exam score of engineering majors equal to​ 88, give
Orlov [11]

Answer: 0.9332.

Step-by-step explanation:

Claim : College Algebra final exam score of engineering majors equal to​ 88.

Given that : The  test statistic is z equals to 1.50.

To find the  p​-value (Probability value), we use standard normal distribution table, and search the p-value corresponds to the z-score.

In a Standard Normal Distribution Table below, the p-value corresponds z equals 1.5 is 0.9332.

Hence, the p​-value is  0.9332.

8 0
3 years ago
NO LINKS!!
dexar [7]

Answer: Anything between 0 and 10, excluding both endpoints.

In terms of symbols we can say 0 < w < 10 where w is the width.

===================================================

Explanation:

You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.

w = width

2w = twice the width

2w-5 = five less than twice the width = length

So,

  • width = w
  • length = 2w-5

which lead to

area = length*width

area = (2w-5)*w

area = 2w^2-5w

area < 150

2w^2 - 5w < 150

2w^2 - 5w - 150 < 0

To solve this inequality, we will solve the equation 2w^2-5w-150 = 0

Use the quadratic formula. Plug in a = 2, b = -5, c = -150

w = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\w = \frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-150)}}{2(2)}\\\\w = \frac{5\pm\sqrt{1225}}{4}\\\\w = \frac{5\pm35}{4}\\\\w = \frac{5+35}{4} \ \text{ or } \ w = \frac{5-35}{4}\\\\w = \frac{40}{4} \ \text{ or } \ w = \frac{-30}{4}\\\\w = 10 \ \text{ or } \ w = -7.5\\\\

Ignore the negative solution as it makes no sense to have a negative width.

The only practical root is w = 10.

If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.

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

Based on that root, we need to try a sample value that is to the left of it.

Let's say we try w = 5.

2w^2 - 5w < 150

2*5^2 - 5*5 < 150

25 < 150 ... which is true

This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.

Now try something to the right of 10. I'll pick w = 15

2w^2 - 5w < 150

2*15^2 - 5*15 < 150

375 < 150 ... which is false

It means w > 10 leads to 2w^2-5w < 150  being false.

Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.

4 0
2 years ago
Can someone please answer these for me?!
Advocard [28]

Answer:

base : 9

three points : (1,0), (9,1), (81,2)

domain : x>0

range : all real number

asymptote : x=0

7 0
3 years ago
Does someone mind helping me with this problem? Thank you!
Jet001 [13]

Answer:

875 ft²

Step-by-step explanation:

Finding area of similar rectangles:

Scale factor = EF : AB

                    = 25 : 5

                    = 5 : 1

\sf \dfrac{Area \ of \ bigger \ rectangle}{Area \ of \ smaller \ rectangle}= (Scale \ factor)^2

\dfrac{Area \ of \ bigger \  rectangle}{35} =\left(\dfrac{5}{1}\right)^2

\text{Area of bigger rectangle = $\dfrac{25}{1}*35$}

Area or rectangle EFGH = 875 ft²

4 0
2 years ago
Read 2 more answers
The vertices of shape K are at (4,7), (7, 7), (7,4) and (5, 5).
son4ous [18]

Answer:

To find the scale factor of the enlargement, compare the distance between a pair of corresponding points from both shapes.

<u>Shape K</u>

A = (4, 7)

B = (7, 7)

C = (7, 4)

D = (5, 5)

Horizontal distance between A (4, 7) and B (7, 7) = 3 units

<u>Shape L</u>

A' = (0, 11)

B' = (9, 11)

C' = (9, 2)

D' = (3, 5)

Horizontal distance between A' (0, 11) and B' (9, 11) = 9 units

9 ÷ 3 = 3

Therefore, Shape L is an enlargement of Shape K by scale factor 3.

To find the center of dilation (enlargement), draw two lines through 2 corresponding points (e.g. A and A',  B and B') - the point of intersection of these lines is the center of dilation.

Therefore, the center of enlargement is (6, 5)  (refer to the second attached image).

6 0
2 years ago
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