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Juliette [100K]
3 years ago
9

I do not understand how to do this problem please help!

Mathematics
2 answers:
Lilit [14]3 years ago
6 0
Please see the attachment for the detailed answer.
I hope it helps.

kicyunya [14]3 years ago
4 0
Y=-4/3x+300 hope this helps
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Factor the equation. Remember the value should add to the b value and multiply to the c value.

x^2-x-12= 0
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Use the zero product property to find roots
x-4=0
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x+3=0
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Final answer: x=-3, x=4
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What's bigger 20 or -25
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Answer:

I hope this helps you

Step-by-step explanation:

20 as it is a positive number

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A company did a quality check on all the packs of trail mixes it manufactured. Each pack of trail mixes is targeted to weigh 9.2
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If each pack of trail mixes is targeted to weigh 9.25 oz and must be within 0.23 oz of the target in order to be accepted, then rejected masses x, are those which weighs less than 9.02 oz or greater than 9.48 oz. 
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I need geometry help
Elanso [62]

Angle BEC is the unmarked angle in the middle of the figure. Its value is the difference between straight angle AED (180°) and the two marked angles, 57° and 33°. That difference is ...

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LT1 10th grade level
Anton [14]

Answer:

The distance between the pirate and the treasure can be found from the following relationship between ΔEAF and ΔABC;

\overline{AE}/\overline{AB} =  \overline{FA}/\overline{AC} = \overline{EF}/\overline{BC}

Step-by-step explanation:

From the question diagram, we have two triangles, ΔEAF and ΔABC;

The ratio of the lengths of the sides \overline{AE}/\overline{AB} =  \overline{FA}/\overline{AC}

∠EAF and ∠BAC are vertical angles, therefore ∠EAF = ∠BAC

Therefore, ΔEAF and ΔABC are similar triangles by Side-Angle-Side, SAS, rule of similarity which states that two triangles that have ratios of a pair of their corresponding sides and the two sides also form equal angles within each triangle, then the two triangles are similar

Therefore, the ratio of the each pair of corresponding sides of the two triangles are equal

We have;

\overline{AE}/\overline{AB} =  \overline{FA}/\overline{AC} = 50 ft. /(100 ft.) =  \overline{EF}/\overline{BC} = \overline{EF}/120 ft.

\overline{EF} = 120 ft. × 50 ft./(100 ft.) = 60 ft.

\overline{EF} = 60 ft.

The distance between the pirate and the treasure, \overline{EF} = 60 ft.

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3 years ago
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