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Paraphin [41]
3 years ago
7

Is domanin x or y???????????????????????????????????

Mathematics
2 answers:
kompoz [17]3 years ago
8 0

Answer:

The domain is x and the Range is y

Mandarinka [93]3 years ago
6 0

Answer:

x-axis

Step-by-step explanation:

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How to solve this problem that don't nor know
Paraphin [41]
Well, for number 23 you have to figure out what n is. In order to do that you have to multiply different numbers to figure out the number for n. I hope this helps and have a wonderful day.
6 0
3 years ago
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Hello need help with this one to and thanks and have a great day or weekend and be safe
Mandarinka [93]

Answer:

a) It's a quadratic

b) f(x) = A(x-3)(x-7)

c) she is right

Step-by-step explanation:

(0-12)/(3-1) = -6

(-4-0)/(5-3) = -2

Since the slope is not constant, it can't be a linear.

Therefore a quadratic

Since the zeroes are 3 and 7,

Factors are (x - 3) and (x - 7)

f(x) = A(x - 3)(x - 7)

Use any one point besides the x-intercepts to find A

12 = A(1-3)(1-7)

12 = 12A

A = 1

f(x) = (x - 7)(x - 3)

f(x) = x² - 7x - 3x + 21

f(x) = x² - 10x + 21

She is right

7 0
3 years ago
Read 2 more answers
The question <br> Najheheggebbsbv nhsygenbs
dezoksy [38]

Answer: 1.) √121= 11

Step-by-step explanation:

11•11=121

4 0
3 years ago
Attachment below <br> algebra helppppp
yKpoI14uk [10]

Answer: second option.

Step-by-step explanation:

In order to solve this exercise, it is necessary to remember the following properties of logarithms:

1)\ ln(p)^m=m*ln(p)\\\\2)\ ln(e)=1

In this case you have the following inequality:

e^x>14

So you need to solve for the variable "x".

The steps to do it are below:

1. You need to apply ln to both sides of the inequality:

ln(e)^x>ln(14)

2. Now you must apply the properties shown before:

(x)ln(e)>ln(14)\\\\(x)(1)>2.63906\\\\x>2.63906

3. Then, rounding to the nearest ten-thousandth, you get:

x>2.6391

3 0
2 years ago
Is the bisector of ∠ABC and is the bisector of ∠ACB.
ycow [4]

Answer:

ASA

Step-by-step explanation:

You can show the angles at either end of segment BC in triangles MBC and LCB are congruent, so you have two angles and the segment between. The appropriate theorem in such a case is ASA.

7 0
3 years ago
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