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vitfil [10]
3 years ago
9

What is the midpoint of the x-intercepts of f(x) = (x – 2)(x – 4)?

Mathematics
2 answers:
tatyana61 [14]3 years ago
8 0
The x-intercepts are readily identified on the graph:  (2,0) and (4,0).  The midpoint of a line connecting those 2 points would be (3,0).
Nutka1998 [239]3 years ago
6 0

Answer:

The correct option is D.

Step-by-step explanation:

The given function is

f(x)=(x-2)(x-4)

From the given graph it is clear that x-intercepts of the graph are (2,0) and (4,0).

The midpoint of x-intercepts is

Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Midpoint=(\frac{2+4}{2},\frac{0+0}{2})

Midpoint=(\frac{6}{2},\frac{0}{2})

Midpoint=(3,0)

The midpoint of x-intercepts is (3,0). Therefore the correct option is D.

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Simplify 4sqrt6/sqrt30 by rationalizing the denominator
andreev551 [17]
4sqrt5/5

see the attachment

8 0
3 years ago
Is the simplified form of 2 square root of 3 ⋅ 2 square root of 6 rational?
alina1380 [7]
√3(√6)
= √3(6)
=√3(3·2)
=3√2

B. No the product will not be rational as 3√2 cannot be expressed as a simple fraction
7 0
3 years ago
Read 2 more answers
Could the equation. For the graph of the function have degree 4?explain
aivan3 [116]

The graphed polynomial seems to have a degree of 2, so the degree can be 4 and not 5.

<h3>Could the graphed function have a degree 4?</h3>

For a polynomial of degree N, we have (N - 1) changes of curvature.

This means that a quadratic function (degree 2) has only one change (like in the graph).

Then for a cubic function (degree 3) there are two, and so on.

So. a polynomial of degree 4 should have 3 changes. Naturally, if the coefficients of the powers 4 and 3 are really small, the function will behave like a quadratic for smaller values of x, but for larger values of x the terms of higher power will affect more, while here we only see that as x grows, the arms of the graph only go upwards (we don't know what happens after).

Then we can write:

y = a*x^4 + c*x^2 + d

That is a polynomial of degree 4, but if we choose x^2 = u

y = a*u^2 + c*u + d

So it is equivalent to a quadratic polynomial.

Then the graph can represent a function of degree 4 (but not 5, as we can't perform the same trick with an odd power).

If you want to learn more about polynomials:

brainly.com/question/4142886

#SPJ1

5 0
2 years ago
Keith has 40-pound bags of mulch in his truck that weigh a total of 3600 pounds. His Owner’s Manual lists the truck’s capacity a
kiruha [24]

Answer:

Keith must remove 90-75 = 15 bags in order to meet the weight requirements

Step-by-step explanation:

This problem can be solved by consecutive rules of three problem.

Rule of three problem:

In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.

When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too. In this case, the rule of three is a cross multiplication

When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease. In this case, the rule of three is a line multiplication.

In this problem, the measures are:

- The number of bags

- The total weight

As the number of bags increases, so do the total weight. So, the relationship between the measures is direct.

To know how many bags does Keith need to remove in order to meet the weight requirements, we first need to know how many bags are in the truck currently.

The problem states that each bag weighs 40 pounds and the weight of the truck is 3600 pounds, so:

1 bag - 40 pounds

x bags - 3600 pounds

40x = 3600

x = \frac{3600}{40}

x = 90.

Keith currently has 90 bags in the truck.

Now, we have to know how much bags he can have in the truck, that is, 3000 pounds worth of bags. So:

1 bag - 40 pounds

x bags - 3000 pounds

40x = 3000

x = \frac{3000}{40}

x = 75.

The truck's capacity is at most 75 bags.

So, Keith must remove 90-75 = 15 bags in order to meet the weight requirements

7 0
3 years ago
Find the missing values in these equivalent fractions 1/2 2/? 3/? 4/?
dalvyx [7]
1/2 = 2/4 = 3/6 = 4/8
6 0
3 years ago
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