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bogdanovich [222]
3 years ago
7

A student scored 81 and 93 on her first two quizzes. Write and solve a compound inequality to

Mathematics
1 answer:
andreev551 [17]3 years ago
6 0
To have an average in between 85 and 90 he has to have a total in-between 255 and 270.her current total is 174 so her third score must be 81≤x≤96

I hope this helps.
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Write and simplify the integral that gives the arc length of the following curve on the given interval b. If necessary, use tech
LiRa [457]

Answer:

L = 4.103

Step-by-step explanation:

we have length of curve

L = \int\limits^b_a {\sqrt{(f'(x))²+1} } \, dx

where f(x) = d/dx(3*in(x)) = 3/x

substituting for f(x), we have L = \int\limits^5_2 {\sqrt{(3/x)²+1} } \, dx

(since the limit is 2≤ x ≤5)

solving,  L = \int\limits^5_2 {\sqrt{9/x²+1} } \, dx

Simplifying this integral, we have

L = 4.10321

8 0
3 years ago
The answer I neeed it now
slava [35]
Anser is B). Subtract smallest number
5 0
3 years ago
Which statements are true about the graph of the function f(x) = x2 – 8x + 5? Check all that apply.
statuscvo [17]

Answer:

A, D, E are true

Step-by-step explanation:

You have to complete the square to prove A.  Do this by first setting the function equal to 0, then moving the 5 to the other side.

x^2-8x=-5

Now we can complete the square.  Take half the linear term, square it, and add it to both sides.  Our linear term is 8 (from the -8x).  Half of 8 is 4, and 4 squared is 16.  So we add 16 to both sides.

(x^2-8x+16)=-5+16

We will do the addition on the right, no big deal.  On the left, however, what we have done in the process of completing the square is to create a perfect square binomial, which gives us the h coordinate of the vertex.  We will rewrite with that perfect square on the left and the addition done on the right,

(x-4)^2=11

Now we will move the 11 back over, which gives us the k coordinate of the vertex.

(x-4)^2-11=y

From this you can see that A is correct.

Also we can see that the vertex of this parabola is (4, -11), which is why B is NOT correct.

The axis of symmetry is also found in the h value.  This is, by definition, a positive x-squared parabola (opens upwards), so its axis of symmetry will be an "x = " equation.  In the case of this type of parabola, that "x = " will always be equal to the h value.  So the axis of symmetry is

x = 4, which is why C is NOT correct, either.

We can find the y-intercept of the function by going back to the standard form of the parabola (NOT the vertex form we found by completing the square) and sub in a 0 for x.  When we do that, and then solve for y, we find that when x = 0, y = 5.  So the y-intercept is (0, 5).

From this you can see that D is also correct.

To determine if the parabola has real solutions (meaning it will go through the x-axis twice), you can plug it into the quadratic formula to find these values of x.  I just plugged the formula into my graphing calculator and graphed it to see that it did, indeed, go through the x-axis twice.  Just so you know, the values of x where the function go through are (.6833752, 0) and (7.3166248, 0).  That's why you need the quadratic formula to find these values.

7 0
3 years ago
Is the function g(x)=9<br> x+1 linear or nonlinear?
Licemer1 [7]
It is a linear function
3 0
3 years ago
(40 pts)can you factor the difference of square?perfect square trinomials?
stira [4]

Answer:

Step-by-step explanation:

The difference of two squares may be represented by the formula:  a^2-b^2,

which can be factored as (a+b)(a-b)

A perfect square trinomial may be represented by the formula: a^(2)-2ab+b^2 or  a^(2)+2ab+b^2, depending on the sign of b

if b is negative: use the formula a^(2)-2ab+b^2, which can be factored as (a-b)*(a-b) or (a-b)^(2)

if b is positive: use the formula a^(2)+2ab+b^2, which can be factored as (a+b)*(a+b) or (a+b)^(2)

7 0
3 years ago
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