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vichka [17]
3 years ago
11

PLEASE HELP !!!!!!!!!!!!!

Mathematics
1 answer:
Akimi4 [234]3 years ago
4 0

9514 1404 393

Answer:

  the multiplicity is 4

Step-by-step explanation:

The graph shows a root at x = -2 that has a multiplicity of 2. You know the multiplicity is even, because the graph does not cross the x-axis. The multiplicity is 2 because the general shape of the graph in that area matches that of a quadratic (parabola).

The multiplicity of the root at x=4 is also an even number, because the x-axis is not crossed. However, the graph is significantly flatter at that point (than at x=-2), meaning the multiplicity is greater than 2. It is at least 4.

When we draw a graph with a multiplicity of 6 at x=4, we find the ratio of the peaks near x=-4 and x=0 to be different from that shown here. The suggests that the multiplicity of the root at x=4 is exactly 4.

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What are the domain, range, and asymptote of h(x) = (0.5)x – 9?
Rzqust [24]

Let's solve for h

hx = 0.5x - 9

Step 1: Divide both sides by x.

hx/x = 0.5x - 9/ x

Answer = h = 0.5x - 9/ x

8 0
2 years ago
The value of n is both 5 times as much as the value of m and 36 more than the value of m. What are the values of n and m? Explai
tia_tia [17]
N = 5m
n = m + 36

since n is equal to both the expression in m we can say:-

5m = m + 36
4m=36

m = 9 

and n = 5*9 = 45

3 0
3 years ago
What is -15x=-240?????????
Vlad [161]
You divide -15 on both sides to keep the x alone
-240/ -15=16
A negative divided by a negative is a positive so x=16
4 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP! I don’t recall how to do this!
MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

(1.05)^x=1.885

In order to solve for x, we have to get it out from its position of an exponent.  Do that by taking the natural log of both sides:

ln(1.05)^x=ln(1.885)

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

x=\frac{ln(1.885)}{ln(1.05)}

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

ln(x*x^2)=5

Simplifying gives you

ln(x^3)=5

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

ln(x)=\frac{5}{3}

Take the inverse ln by raising each side to e:

e^{ln(x)}=e^{\frac{5}{3}}

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

\frac{1}{2}=e^{.1x}

"Undo" that e by taking the ln of both sides:

ln(.5)=ln(e^{.1x})

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

-.6931471806 = .1x

x = -6.931471806

Question d. is a bit more complicated than the others.  Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

(2^2)^x-6(2)^x=-8

Now we will bring over the -8 by adding:

(2^2)^x-6(2)^x+8=0

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution.  Let's let u=2^x

When we do that, we can rewrite the polynomial as

u^2-6u+8=0

This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor.  Now we have to put it back in:

2^x=4,2^x=2

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

2^2=2^x

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

2^x=2^1

Now that the bases are the same, we can say that

x = 1

5 0
2 years ago
90 points? pleaseeee​
Dmitry_Shevchenko [17]

Answer:

y = 28 degrees

x = 28 degrees

Step-by-step explanation:

A right angle = 90 degrees

90 degrees - 62 degrees = 28 degrees

This is for both x and y because 62 degrees occupies both the right angles.

8 0
3 years ago
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