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Elena-2011 [213]
3 years ago
9

Answer answer answer ​

Mathematics
1 answer:
Sergio [31]3 years ago
6 0

Answer:

(m−5)(2m−2)

=(m+−5)(2m+−2)

=(m)(2m)+(m)(−2)+(−5)(2m)+(−5)(−2)

=2m2−2m−10m+10

=2m2−12m+10

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Answer:

C. f(x) = 3·x + 2, g(x) = 7·x + 6

Step-by-step explanation:

The given equations relates to the property of equality of values;

The given formula for the association between f(x) and g(x) is f(x) = g(x)

The given equation of two expressions is 3·x + 2 = 7·x + 6

By transitive property of equality, the two above equations are correct when f(x) = 3·x + 2 and g(x) = 7·x + 6

Therefore, the function that may be used to represent the equation is option C; f(x) = 3·x + 2, g(x) = 7·x + 6.

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Can someone help me plz .. Use these lengths to find cos B, tan B, and sin B.
lara [203]

Answer:

cos B = \frac{7}{25}

tan B = \frac{24}{7}

sin B = \frac{24}{25}

Step-by-step explanation:

In the right triangle, there are three sides and 2 acute angles

  • Hypotenuse ⇒ the opposite side of the right angle
  • Leg1 and Leg 2 ⇒ the sides of the right angle

The trigonometry functions of one of the acute angles Ф are

  • sin Ф = opposite leg/hypotenuse
  • cos Ф = adjacent leg/hypotenuse
  • tan Ф = opposite leg/adjacent leg

In Δ ACB

∵ ∠C is the right angle

∴ AB is the hypotenuse

∵ AC is the opposite side of ∠B ⇒ leg1

∵ CB is the adjacent side of ∠B ⇒ leg2

→ By using the ratios above

∴ cos B = \frac{CB}{AB} , tan B = \frac{AC}{CB} , sin B = \frac{AC}{AB}

∵ CB = 7, AB = 25, AC = 24

∴ cos B = \frac{7}{25}

∴ tan B = \frac{24}{7}

∴ sin B = \frac{24}{25}

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write and graph the equation of a line passing through point 1,3 that is parallel and perpendicular to the line with the equatio
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Answer: y=2x+1 is parallel & y=-2x+5 is perpendicular



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