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kupik [55]
3 years ago
6

HELP!!! Find the length of side x in simplest radical form with a rational denominator.

Mathematics
1 answer:
Leokris [45]3 years ago
7 0

Answer:

Step-by-step explanation:

Square root of 2

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Write an equation in point slope form of a line passing through 3,6 having a slope of 1/3
Rina8888 [55]

Answer:

Step-by-step explanation:

slope (m) = 1/3

y - intercept (c) = 6

writing in point slope form

y = mx + c

y = 1/3 x + 6

8 0
3 years ago
Only answer if you're very good at math.Please I keep on posting this but nobody is helping me.
hichkok12 [17]
I believe A. but Major guess tho
3 0
3 years ago
For the following exercises, determine whether each of the following relations is a function.
RSB [31]

Answer:

The relation {(2,1),(3,2),(-1,1),(0,-2)} is a function because the domain {2,3,-1,0} and range {1,2,1,-2} of the relation is paired exactly only once.

Step-by-step explanation:

The given relation is {(2,1),(3,2),(-1,1),(0,-2)}.

It is required to determine the relation is a function.

Step 1 of 1

The given relation is {(2,1),(3,2),(-1,1),(0,-2)}.

This relation shows that the domain of the relation {2,3,-1,0} is paired with one element of the range {1,2,1,-2}.

Hence, the relation is a function.

7 0
2 years ago
What is the solution set to this equation?<br> log_4(x + 3) + log_4x = 1
sesenic [268]

Answer:

x=1

Step-by-step explanation:

log_4(x + 3) + log_4x = 1

We know that loga(b) + loga(c) = loga(bc)

log_4(x + 3)x = 1

Raise each side to the base of 4

4^log_4(x + 3)x = 4^1

(x+3)x = 4

x^2 +3x = 4

Subtract 4 from each side

x^2 +3x -4 = 0

Factor

(x+4) (x-1) =0

Using the zero product property

x= -4  x=1

But x cannot be negative since logs cannot be negative

x=1

6 0
3 years ago
Read 2 more answers
Which point is on the graph of the function
natulia [17]

Answer:

<h2>(1, 1)</h2>

Step-by-step explanation:

(x,\ y)\\\\\text{Put thecoordinates of the points to the equation of}\ f(x)\ \text{and check the equality}:\\f(x)=\dfrac{1}{2}(2)^x\\\\(0,\ 1)\to x=0,\ y=1\\L=1\\R=\dfrac{1}{2}(2)^0=\dfrac{1}{2}(1)=\dfrac{1}{2}\\\\L\neq R\\\\(0,\ 2)\to x=0,\ y=2\\L=2\\R=\dfrac{1}{2}\\\\L\neq R\\\\\left(1;\ \dfrac{1}{2}\right)\to x=1,\ y=\dfrac{1}{2}\\L=\dfrac{1}{2}\\R=\dfrac{1}{2}(2)^1=\dfrac{1}{2}(2)=1\\\\L\neq R\\\\(1,\ 1)\to x=1,\ y=1\\L=1\\R=1\\\\L=R

4 0
3 years ago
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