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lozanna [386]
3 years ago
7

Through:(4,0) slope =1/2

Mathematics
2 answers:
abruzzese [7]3 years ago
4 0

Answer: M = - 2/3

Step-by-step explanation:

Not sure this should be the answer tho

kakasveta [241]3 years ago
4 0

Answer:

M = - 2/3 Not sure.

Step-by-step explanation:

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Amiraneli [1.4K]

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8 0
3 years ago
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Find the most common ratio 72,12,2, 1/3,1/18
liraira [26]

Answer:

1/6

Step-by-step explanation:

To find the common ratio, you compare a few pairs of consecutive terms, by dividing an element by its predecessor.

12 / 72 = 1/6

2 / 12 = 1 / 6

1/3 / 2 = 1 / 6

The ratio is constant... so that's your common ratio to go from one term to the next.  

To go from one term to the next, you have to multiply by 1/6.

7 0
3 years ago
Please help! Explain to me how you found your answer.
Katena32 [7]

Step-by-step explanation:

Sin: 2 square root 3: 4

Cos: 2:4

Tan: 2 square root 3: 2

6 0
3 years ago
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An acute triangle has sides measuring 10 cm and 16 cm. The length of the third side is unknown.
saveliy_v [14]

Answer: Choice B

12.5 < x < 18.9

================================================

Explanation:

We have a triangle with these side lengths:

  • a = 10
  • b = 16
  • c = x = unknown

Let's assume that b = 16 is the largest side of this triangle.

By the converse of the pythagorean theorem, we need b^2 < a^2+c^2 to be true in order for an acute triangle to happen.

So,

b^2 < a^2 + c^2\\\\c^2 > b^2 - a^2\\\\c > \sqrt{b^2-a^2}\\\\x > \sqrt{16^2-10^2}\\\\x > \sqrt{156}\\\\x > 12.4899959967968 \ \text{(approximate)}\\\\x > 12.5

Now let's consider the possibility that the missing side x is actually the longest side.

Using the same theorem as before, we would say,

c^2 < a^2 + b^2\\\\c < \sqrt{a^2 + b^2}\\\\x < \sqrt{10^2 + 16^2}\\\\x < \sqrt{356}\\\\x < 18.8679622641132 \ \text{(approximate)}\\\\x < 18.9\\\\

We found that x > 12.5 and x < 18.9

This is the same as saying 12.5 < x and x < 18.9

Put together, they form the approximate answer of 12.5 < x < 18.9

6 0
3 years ago
Which expression is the factorization of x2 + 10x + 21?
LiRa [457]
<h2><u>Sol</u><u>ution</u><u>:</u></h2>

Equation: x² + 10x + 21

<u>Step</u><u> </u><u>1</u><u>:</u> Find two numbers that can add up to 10 and be multiplied to 21. We have: 7 & 3, in the sense that 7+3=10, and 7×3=21. Replacing 10 with 7+3, the equation is now → x² + 7x + 3x + 21

<u>Step</u><u> </u><u>2</u><u>:</u> Get the new equation bracketed → (x² + 7x) (+3x + 21)

<u>Step</u><u> </u><u>3</u><u>:</u> Use 'x' in the equation. For the first part, we have 'x'. x² = x × x so, bring out one x out side the bracket, divide 7x by = 7 → x (x +7). Do the same for the second part by dividing 21 by 3 = 7, and then bringing out 3 from the bracket → 3 (x + 7).

Bringing everything together, we have: x(x+7) +3(x+7) → (x+3) (x+7)

<h3><u>Final</u><u> </u><u>ans</u><u>wer</u><u>:</u></h3>

(x+3) (x+7)

<h3 />

4 0
2 years ago
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