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MArishka [77]
4 years ago
14

Solve for x. x/4 = -1​

Mathematics
2 answers:
Lyrx [107]4 years ago
8 0

Answer: -4

Step-by-step explanation: -4/4 = -1

Please mark Brainliest!

Yakvenalex [24]4 years ago
8 0

Answer:

x = -4

Step-by-step explanation:

x / 4 = -1

multiply both sides by 4

which gets you

x = -4

hope this helps

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A cookie company packages its cookies in rectangular prism boxes designed with square bases that have both a length and width of
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Answer:

h³- 8h² + 16h

Step-by-step explanation:

The problem tells us that the length and width of these boxes are both 4 inches less than the height of the box.

So if we name <u>h the height of the box</u>, the <u>width of the box would be h - 4 </u>and the <u>height of the box would be h - 4.</u>

Now, the volume  of a rectangular prism is given by V = height x width x length

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Therefore, the polynomial representing the volume of this box in terms of the height is h³- 8h² + 16h

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3 years ago
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sergij07 [2.7K]

Answer:

I didn't know if you wanted to get the answer

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3 years ago
Gail Leonard�s monthly take home pay is $5,000 She has a $600 mortgage payment, $750 car payment, $120 student loan payment, and
storchak [24]
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Can someone help me on this pls? It’s urgent, so ASAP (it’s geometry)
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<u>Question 6</u>

1) \overline{AB} \cong \overline{BD}, \overline{CD} \perp \overline{BD}, O is the midpoint of \overline{BD}, \overline{AB} \cong \overline{CD} (given)

2) \angle ABO, \angle ODC are right angles (perpendicular lines form right angles)

3) \triangle ABO, \triangle CDO are right triangles (a triangle with a right angle is a right triangle)

4) \overline{BO} \cong \overline{OD} (a midpoint splits a segment into two congruent parts)

5) \triangle ABO \cong \triangle CDO (LL)

<u>Question 7</u>

1) \angle ADC, \angle BDC are right angles), \overline{AD} \cong \overline{BD}

2) \overline{CD} \cong \overline{CD} (reflexive property)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \triangle ADC \cong \triangle BDC (LL)

5) \overline{AC} \cong \overline{BC} (CPCTC)

<u>Question 8</u>

1) \overline{CD} \perp \overline{AB}, point D bisects \overline{AB} (given)

2) \angle CDA, \angle CDB are right angles (perpendicular lines form right angles)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \overline{AD} \cong \overline{DB} (definition of a bisector)

5) \overline{CD} \cong \overline{CD} (reflexive property)

6)  \triangle ADC \cong \triangle BDC (LL)

7) \angle ACD \cong \angle BCD (CPCTC)

8 0
2 years ago
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