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seraphim [82]
3 years ago
9

Simplify -8 (2-3w) - 5w

Mathematics
2 answers:
Vika [28.1K]3 years ago
7 0
The answer is -16 + 19w
First distribute -8, then collect like terms and simplify.
-16 + 24w - 5w
=-16 + 19w
nika2105 [10]3 years ago
5 0

<em>Answer,</em>

<em><u>= -16 + 19w</u></em>

<em>Explanation,</em>

<em><u>Distribute:</u></em>

<em>= (−8)(2) + (−8)(−3w) + −5w</em>

<em>= −16 + 24w + −5w</em>

<em><u>Combine Like Terms:</u></em>

<em>= −16 + 24w + −5w = (24w + −5w) + (−16) </em>

<em><u>= 19w + −16</u></em>


<u><em>Hope This Helps :-)</em></u>

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3 0
3 years ago
Solve the equation 29 - (x+8) = 6x - 7
max2010maxim [7]
First step you should do is s<span>implify both sides of your equation:
</span>29-(x+8)=6x-7
Distribute the Negative Sign:
29+-1(x+8)=6x-7
29+-1x+(-1)(8)=6x-7
29+-x+-8=6x-7
29+-x+-8=6x+-7
<span>Combine Like Terms:
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</span>-7x/-7 = -28/-7
And now your answer should be:
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6 0
3 years ago
7. What is the force of tension in a rope that spins a 250 g object in a vertical circle with a radius of 50 cm if the
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4 0
3 years ago
1.Find the zeros of f(x) = x2 - 9
Marat540 [252]
So for the first one it is (x-3)(x+3) this is because it is a difference of squares so you square root the 9 and the minus you do both +and - in order to keep up the -9
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6 0
3 years ago
In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th
olya-2409 [2.1K]

Answer:

\frac{BE}{EC} =\frac{1}{3}

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

\frac{BK}{CK} =\frac{1}{4}

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE                ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE     ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK  and Δ EBK are similar

so we have

\frac{AD}{BE} =\frac{DK}{BK}

\frac{AD}{BE} =\frac{4}{1}

\frac{BC}{BE}=\frac{4}{1}     ( ABCD is parallelogram so AD=BC)

\frac{BE+EC}{BE}=\frac{4}{1}         ( BC= BE+EC)

\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}

1+\frac{EC}{BE}=4

\frac{EC}{BE}=3  ( subtracting 1 from both side )

\frac{EC}{BE}=\frac{3}{1}

taking reciprocal both side

\frac{BE}{EC} =\frac{1}{3}


8 0
3 years ago
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