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Fofino [41]
3 years ago
9

Determine the solution set of x^2-80=0

Mathematics
1 answer:
Goryan [66]3 years ago
5 0
<span>x^2-80=0
</span><span>x^2 = 80
x^2 = 16 (5)
x = 4</span>√5 and x = -4<span>√5</span>
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Which set of ordered pairs in the form of (x,y) does not represent a function of x? {(-1,2), (3,-2),(0,1),(5,2}
serg [7]
For a relation to be a function it must be one-to-one or many-to-one.

One-to-many relation is not a function.

In the above options, C is One-to-many relation therefore cannot be a function.

The reason is that 3 alone maps onto -2 and 5, in the ordered pairs (3,-2) and (3,5). This disqualifies it from being a function.

Hence the graph of this relation will not pass the vertical line test

The correct answer is C
5 0
3 years ago
I need help it’s for a very hard test
Mumz [18]

Answer:

C

Step-by-step explanation

This has a factor of 6% or 0.06. But, you are adding and not subtracting so it is 1.06. ( If you multiply by a number less than 1 then it is actually dividing) So C is the only on that is correct.

6 0
3 years ago
Use the information given in the diagram. Tell why AB≈DC and
Katen [24]
AB=DC because they have the same side lengths
5 0
3 years ago
Read 2 more answers
Help with 21 would be much appreciated.
umka2103 [35]

Answer:

\large\boxed{x=2\sqrt{21}\ and\ y=4\sqrt3}

Step-by-step explanation:

Look at the picture.

ΔADC and ΔCDB are similar. Therefore the sides are in proportion:

\dfrac{AD}{CD}=\dfrac{CD}{DB}

We have

AD=14-8=6\\CD=y\\DB=8

Substitute:

\dfrac{6}{y}=\dfrac{y}{8}         <em>cross multiply</em>

y^2=(6)(8)

y^2=48\to y=\sqrt{48}\\\\y=\sqrt{16\cdot3}\\\\y=\sqrt{16}\ cdot\sqrt3\\\\\boxed{y=4\sqrt3}

For x use the Pythagorean theorem:

x^2=6^2+(4\sqrt3)^2\\\\x^2=36+48\\\\x^2=84\to x=\sqrt{84}\\\\x=\sqrt{4\cdot21}\\\\x=\sqrt4\cdot\sqrt{21}\\\\\boxed{x=2\sqrt{21}}

3 0
3 years ago
How do I solve this problem using the substitution method?
Ira Lisetskai [31]
<span>(y=mx+b) or (ax+by=c)  hope this helped

</span>
8 0
3 years ago
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