If the graph of a line goes from the upper left corner of the graph to the lower right corner, what can you say about its slope?

Mathematics

1 answer:

musickatia [10]2 years ago

4 0

Answer:

I believe the answer is D. Hope this helps

Step-by-step explanation:

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Evaluate this exponential expression.

Afina-wow [57]

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2 years ago

Side legnths for x^2 -4x -12

Ugo [173]

I hope this helps you

x^2-4x-12

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x +4

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3 years ago

Prove the following relations: 1) (1 - cos²A) (1 + tan²A) = tan²A

liberstina [14]

Step-by-step explanation:

(1 - cos²A) (1 + tan²A) = tan²A

(sin²A)(sec²A) = tan²A

(sin²A)(1/cos²A) = tan²A

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2 years ago

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Factor the following polynomial completely. 6x^5-30x^4-84x^3

Lerok [7]

6x^5 - 30x^4 - 84x^3

we can factor out 6x^3

6x^3(x^2 - 5x - 14)

Also, we can factor x - 7 and x + 2 out of x^2 - 5x - 14

The final answer = 6x^3(x - 7)(x + 2)

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2 years ago

NEED ANSWER ASAP REALLY IMPORTANT i will take any of the answers

dedylja [7]

$1. \: MC \: is \: tangent \: to \: (B)\Rightarrow \widehat{BCM}=90° \\ 2. \: MC \: and \: MZ \: are \: tangent \: to \: (B)\Rightarrow MC = MZ\Leftrightarrow 5x - 9 = x + 7\Leftrightarrow x = 4 \\ 3. \: BM = \sqrt{ {BC}^{2} + {MC}^{2} } = \sqrt{ {5}^{2} + {12}^{2} } = 13 \\ \Rightarrow EM = BM - BE = BM - BC = 13 - 5 = 8 \\ 4. \: \tan(60°) = \frac{MC}{BC} = \frac{13 \sqrt{3} }{ BC}\Leftrightarrow BC = \frac{13 \sqrt{3} }{\tan(60°)} = \frac{13 \sqrt{3} }{ \sqrt{3} } = 13 \\ 5. \: MC = \sqrt{ {BM}^{2} - {BC}^{2} } = \sqrt{ {20}^{2} - {12}^{2} } = 16 \\ 6. \:BZ = \sqrt{ {BM}^{2} - {MZ}^{2} } = \sqrt{ {25}^{2} - {20}^{2} } = 15 \\ \Rightarrow XE=BX+BE=BZ+BZ=15+15=30$

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3 years ago