What is the quotient of 28 divided by 7 increased by 4

Answers

Answer 1
Answer:

The quotient after increased by 4 is 8.

To calculate the quotient of 28 divided by 7, we perform the division and get:

28 ÷ 7

= 28/7

= (7 x 4)/ 7

= 4

Next, we are asked to increase this quotient by 4. Adding 4 to the quotient:

= 4 + 4

= 8.

Therefore, the quotient after increased by 4 is 8.

Learn more about Division here:

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Answer 2
Answer:

Answer:

8 beacaus the seven increased from 4


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5-11. A polymer is manufactured in a batch chemical process. Viscosity measurements are normally made on each batch, and long experience with the process has indicated that the variability in the process is fairly stable with 20. Fifteen batch viscosity measurements are given as follows: 724, 718, 776, 760, 745, 759, 795, 756, 742, 740, 761, 749, 739, 747, 742. A process change is made that involves switching the type of catalyst used in the process. Following the process change, eight batch viscosity measurements are taken: 735, 775, 729, 755, 783, 760, 738, 780. Assume that process variability is unaffected by the catalyst change. Find a 90% CI on the difference in mean batch viscosity resulting from the process change. What is the practical meaning of this interval

Answers

Step-by-step explanation:

Confidence interval = 90% = 0.90.

1 - 0.90 = 0.10

Mean of 15 batch:

724+718+776+760+745+759+795+756+742+740+761+749+739+747+742 = 11253/15

= 750.2

Mean of second batch:

735+775+729+755+783+760+738+780

= 6055/8

= 756.875

Sd = 20

Z-alpha/2 = 1.64

= (750.2-756.875)-1.64*√20²/15 + 20²/8

= -21.0348

(750.2-756.875)+1.64*√20²/15 + 20²/8

= 7.68

2. The upper limit of the interval is less than 10 so in conclusion the difference in mean batch viscosity is about 10 or less than 10

The integral \pi\int\limits^2_1 {(1-(lnx)^(2)) } \, dx represents the volume of a solid obtained by rotating a region around y=-1. Evaluate.

Answers

Answer:

V = π (-2 (ln 2)² + 4 ln 2 − 1)

V ≈ 2.55

Step-by-step explanation:

V = π ∫₁² (1 − (ln x)²) dx

V/π = ∫₁² (1 − (ln x)²) dx

V/π = ∫₁² dx − ∫₁² (ln x)² dx

V/π = x |₁² − ∫₁² (ln x)² dx

V/π = 1 − ∫₁² (ln x)² dx

To evaluate the second integral, integrate by parts.

If u = (ln x)², then du = 2 (ln x) / x dx.

If dv = dx, then v = x.

∫ u dv = uv − ∫ v du

= (ln x)² x − ∫ x (2 (ln x) / x) dx

= x (ln x)² − 2 ∫ ln x dx

Integrate by parts again.

If u = ln x, then du = 1/x dx.

If dv = dx, then v = x.

∫ u dv = uv − ∫ v du

= x ln x − ∫ x (1/x dx)

= x ln x − ∫ dx

= x ln x − x

Substitute:

∫ (ln x)² dx = x (ln x)² − 2 ∫ ln x dx

∫ (ln x)² dx = x (ln x)² − 2 (x ln x − x)

∫ (ln x)² dx = x (ln x)² − 2x ln x + 2x

Substitute again:

V/π = 1 − ∫₁² (ln x)² dx

V/π = 1 − (x (ln x)² − 2x ln x + 2x) |₁²

V/π = 1 + (-x (ln x)² + 2x ln x − 2x) |₁²

V/π = 1 + (-2 (ln 2)² + 4 ln 2 − 4) − (-1 (ln 1)² + 2 ln 1 − 2)

V/π = 1 − 2 (ln 2)² + 4 ln 2 − 4 + 2

V/π = -2 (ln 2)² + 4 ln 2 − 1

V = π (-2 (ln 2)² + 4 ln 2 − 1)

V ≈ 2.55

What is the least common denominator of the equation StartFraction 2 Over 9 EndFraction x + two-thirds x = 7?

Answers

Answer:

the answer is b

Step-by-step explanation:

Answer: b

Step-by-step explanation:

Simplify 12x7y3 divided by 6x3y

Answers

Answer:

12x7y3

----------- = 6x4y3

 6x3y

Step-by-step explanation:

In dividing these numbers, you simply subtract 6x from 12x and subtract 3y from 7y because 6x and 3y are both in the denominator. 3 is left alone.

Which of the following statements could he use to prove his claim? select all that apply.

Answers

9514 1404 393

Answer:

  B, D, E

Step-by-step explanation:

Any of the following will put rectangle 1 on top of rectangle 2:

  • reflection over the x-axis, translation right 7 units (B)
  • rotation 180° about the origin, translation right 1 unit (D)
  • reflection over the y-axis, translation right 1 and up 6 units (E)

Which graph represents the function f(x) = |x + 3|?

Answers

Answer:

Answer is B

Step-by-step explanation:

If you are unsure about where to start, you could always plot some numbers down until you see a general pattern.

But a more intuitive way is to determine what happens during each transformation.

A regular y = |x| will have its vertex at the origin, because nothing is changed for a y = |x| graph. We have a ray that is reflected at the origin about the y-axis.

Now, let's explore the different transformations for an absolute value graph by taking a y = |x + h| graph.

What happens to the graph?

Well, we have shifted the graph -h units, just like a normal trigonometric, linear, or even parabolic graph. That is, we have shifted the graph h units to its negative side (to the left).

What about the y = |x| + h graph?

Well, like a parabola, we shift it h units upwards, and if h is negative, we shift it h units downwards.

So, if you understand what each transformation does, then you would be able to identify the changes in the shape's location.

Answer:

answer is b

Step-by-step explanation:

because i got it right on E D G E