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OLga [1]
3 years ago
7

1)

Mathematics
2 answers:
SIZIF [17.4K]3 years ago
6 0

Answer: I think A

Step-by-step explanation:

lakkis [162]3 years ago
5 0
It’s D because the original is at +1 and it goes down to -1. If you count backwards starting at 1 to -1 you notice that it’s two digits apart
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How many extraneous solutions does the equation below have?
FrozenT [24]

<u>Answer</u>

0


<u>Explanation</u>

(2m)/(2m+3)-(2m)/(2m-3)=1

Simplifying the left hand first

(2m)/(2m+3)-(2m)/(2m-3) = {2m(2m-3) - 2m(2m+3)}/(4m²-9)

                                        (4m²-6m-4m²-6m)/(4m²-9)

                                         = (-12m) / (4m²-9)

Now this equet to 1

(-12m) / (4m²-9) = 1

-12m =  4m²-9

4m²+ 12m -9 =0 ⇒⇒⇒ This is a quadratic equation that has 2 real solutions.

4m²+ 12m -9 =0

m² + 3m + (3/2)²= 9/4 + 9/4

(m + 3/2)² = 18/4

m = √18/2 - 3/2                or m = -√18/2 - 3/2

    = 0.621                                       = -3.621

So we can say that the equation has NO extraneous solutions.

Answer = 0

8 0
3 years ago
Read 2 more answers
URGENT please help me with this !!!
marin [14]

Answer:

480 mm³

Step-by-step explanation:

The volume (V) of a pyramid is

V = \frac{1}{3} area of base × height, hence

V = \frac{1}{3} × 96 × 15 = 32 × 15 = 480

5 0
3 years ago
Please answer I will give brainliest to the best
IgorC [24]
Answer should be A

Explanation: i’m learning it rn and Experimental probability is the result of an experiment. Theoretical probability is what is expected to happen.
8 0
3 years ago
How can you write the expression with rationalized denominator? 2+sqrt3(3)/sqrt3(6)
joja [24]
So we have a 6 at the bottom, and the root is 3, so hmm how to take it out, simple enough, just let's get something to make the 6 a 6³, so it comes out of the root

so 

\bf \cfrac{2+\sqrt[3]{3}}{\sqrt[3]{6}}\cdot \cfrac{\sqrt[3]{6^2}}{\sqrt[3]{6^2}}\implies \cfrac{(2+\sqrt[3]{3})(\sqrt[3]{6^2})}{(\sqrt[3]{6})(\sqrt[3]{6^2})}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3}\cdot \sqrt[3]{36}}{\sqrt[3]{6^3}}&#10;\\\\\\&#10;\cfrac{2\sqrt[3]{36}+\sqrt[3]{3\cdot 36}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{108}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3^3\cdot 4}}{6}&#10;\\\\\\&#10;\cfrac{2\sqrt[3]{36}+3\sqrt[3]{ 4}}{6}
8 0
3 years ago
PLS HELP 10 brain list
loris [4]

Answer:

y:))))))))))))))))))))

6 0
3 years ago
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