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Black_prince [1.1K]
3 years ago
12

I NEED HELP ASAP Which of the following possibilities will form a triangle

Mathematics
1 answer:
Sholpan [36]3 years ago
8 0

Answer:

The answer is D!

Step-by-step explanation:

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A die is rolled. The set of equally likely outcomes is {1, 2, 3, 4, 5, 6}. Find the probability of getting a 9.
mrs_skeptik [129]

Answer:

you are only rolling 1 dice the probability of getting 9 is 0 there are only 6 sides to one die

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3 years ago
How do I Simplify (5x-2)4 using the distributive property?
BartSMP [9]

Answer:

20x - 8

Step-by-step explanation:

using the distributive property to simplify (5x-2)4, you would multiply the outside term (4) by each inside term (5x-2).

for example:

4 × 5x = 20x

4 × -2 = -8

once you distribute the 4 into 5x-2, you are left with 20x - 8 which needs no further simplifying

4 0
3 years ago
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Pls only do this if yk it because i’m giving correct answer brainliest! :))
Leno4ka [110]

Answer:

x=2+\frac{1}{2}\sqrt[]{21}

or

x=2-\frac{1}{2}\sqrt{{21}

Step-by-step explanation:

4x^2-16x-26=-21

Add 21 on both sides.

4x^2-16x-26+21=-21+21

4x^2-16x-5=0

a=4

b=-16

c=-5

x=\frac{-b\frac{+}{}\sqrt[]{b^2-4ac}  }{2a}

x=\frac{-(-16)\frac{+}{}\sqrt[]{(-16)^2-4(4)(-5)}  }{2(4)}

x=\frac{16\frac{+}{}\sqrt[]{256+80}  }{8}

x=\frac{16\frac{+}{}\sqrt[]{336}  }{8}

x=\frac{16\frac{+}{}\sqrt[]{2^2*2^2*21}  }{8}

x=\frac{16\frac{+}{}2*2\sqrt[]{21}  }{8}

x=\frac{16\frac{+}{}4\sqrt[]{21}  }{8}

---------------------------------------------------------------------------

x=\frac{16}{8}+\frac{4\sqrt[]{21}}{8}

x=2+\frac{1}{2}\sqrt[]{21}

---------------------------------------------------------------------------

x=\frac{16}{8}-\frac{4\sqrt[]{21}}{8}\\x=2-\frac{1}{2}\sqrt{{21}

6 0
3 years ago
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Mike and his 3 friends were eating pizza. They had 2 pizzas that they were going to
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Each person would get 1 and left overs r in the trash lel

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3 years ago
A group of which of the order must be cyclic?<br><br> A. 6 <br> B. 7 <br> C. 8 <br> D. 9
DochEvi [55]

Answer:

Its option D. 9

mark me as brainlist

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