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3 years ago

Simplify: 2b - 15d - 8b + 10b + 12d=

Mathematics

1 answer:

guapka [62]3 years ago

5 0

Answer:

4b-3d

Step-by-step explanation:

first add the like terms

2b - 8b + 10b = 4b

12d - 15d = -3d

then add the results.

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just olya [345]

Answer:

${( \sqrt{ {x}^{ - 3} }) }^{5} = ({( {x}^{ - 3}) }^{ \frac{1}{2} } ) ^{5} \\ \\ = {x}^{ - 3 \times \frac{1}{2} \times 5} = {x}^{ - \frac{15}{2} } = \frac{1}{ {x}^{ \frac{15}{2} } }$

<em><u>Or ;</u></em>

${( \sqrt{ {x}^{ - 3} } )}^{5} = {( \sqrt{ \frac{1}{ {x}^{3} }} })^{5} = ( { \frac{ \sqrt{1} }{ \sqrt{ {x}^{3} } } })^{5} \\ \\ = ( { \frac{1}{ \sqrt{x} \sqrt{ {x}^{2} } } })^{5} = ({ \frac{1}{ x\sqrt{x} }})^{5} \\ \\ = ( \frac{ {(1)}^{5} }{ {x}^{5} ( \sqrt{x} )^{5} } ) = \frac{1}{ {x}^{5} \times {x}^{ \frac{1}{2} \times 5 } } \\ \\ = \frac{1}{ {x}^{5} \times {x}^{ \frac{5}{2} } } = \frac{1}{ {x}^{5} \sqrt{ {x}^{5} } } \\ \\ = \frac{1}{ {x}^{5} \sqrt{ {x}^{4} {x}^{1} } } = \frac{1}{ {x}^{5} \sqrt{x} \sqrt{( { {x}^{2} })^{2} } } \\ \\ = \frac{1}{ {x}^{5} {x}^{2} \sqrt{x} } = \frac{1}{ {x}^{7} \sqrt{x} } = \frac{ \sqrt{x} }{ {x}^{8} }$

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1 year ago

In triangle △ABC, ∠ABC=90°, BH is an altitude. Find the missing lengths. AC=5 and BH=2, find AH and CH.

Yakvenalex [24]

Answer:

AH = 1 or 4

CH = 4 or 1

Step-by-step explanation:

An altitude divides a right triangle into similar triangles. That means the sides are in proportion, so ...

AH/BH = BH/CH

AH·CH = BH²

The problem statement tells us AH + CH = AC = 5, so we can write

AH·(5 -AH) = BH²

AH·(5 -AH) = 2² = 4

This gives us the quadratic ...

AH² -5AH +4 = 0 . . . . in standard form

(AH -4)(AH -1) = 0 . . . . factored

This equation has solutions AH = 1 or 4, the values of AH that make the factors be zero. Then CH = 5-AH = 4 or 1.

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2 years ago

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Dvinal [7]

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