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alexgriva [62]
3 years ago
5

Write 9 x 10^-2 in standard notation.

Mathematics
1 answer:
blondinia [14]3 years ago
4 0

Hello! Your answer is shown below,

= 9 × 10-₂

= 9e-2

(scientific e notation)

= 90 × 10-₃

(engineering notation)

(thousandth; prefix milli- (m))

= 0.09

(real number)

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the length of a rectangle is 5 units longer than its width. If the perimeter must be at least 34 units, what are the possible le
ololo11 [35]

Answer:

w ≥ 6

l ≥ 11

Step-by-step explanation:

The perimeter of a rectangle is equal to P = 2l+2w where l=length and w=width. Here the length is 5 feet longer than the width or 5+w. This means the width is w. Substitute P = 34, l=5+w and w into the perimeter equation. Then solve for w.

34 ≤ 2*(5+w) + 2w

34 ≤ 10+2w+2w

34 ≤ 10+4w

24 ≤ 4w

6 ≤ w

This means the width must be at least 6 so the solution is w ≥ 6.

To find the length substitute 6 into l ≤ 5+w.

l ≤ 5 + 6

l ≤ 11

The length is l ≥ 11.

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Which situation is represented by the equation x+10=26?
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Answer:

16

Step-by-step explanation:

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Factor 5(2x+7)+4(2x+7)<br>enter your answer in the box​
AfilCa [17]

Answer:

18x+63

Step-by-step explanation:

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Combine Like Terms:

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<u>Question 4</u>

1) \overline{BD} bisects \angle ABC, \overline{EF} \perp \overline{AB}, and \overline{EG} \perp \overline{BC} (given)

2) \angle FBE \cong \angle GBE (an angle bisector splits an angle into two congruent parts)

3) \angle BFE and \angle BGE are right angles (perpendicular lines form right angles)

4) \triangle BFE and \triangle BGE are right triangles (a triangle with a right angle is a right triangle)

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

6) \triangle BFE \cong \triangle BGE (HA)

<u>Question 5</u>

1) \angle AXO and \angle BYO are right angles, \angle A \cong \angle B, O is the midpoint of \overline{AB} (given)

2) \triangle AXO and \triangle BYO are right triangles (a triangle with a right angle is a right triangle)

3) \overline{AO} \cong \overline{OB} (a midpoint splits a segment into two congruent parts)

4) \triangle AXO \cong \triangle BYO (HA)

5) \overline{OX} \cong  \overline{OY} (CPCTC)

<u>Question 6</u>

1) \angle B and \angle D are right angles, \overline{AC} bisects \angle BAD (given)

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

3) \angle BAC \cong \angle CAD (an angle bisector splits an angle into two congruent parts)

4) \triangle BAC and \triangle CAD are right triangles (a triangle with a right angle is a right triangle)

5) \triangle BAC \cong \triangle DCA (HA)

6) \angle BCA \cong \angle DCA (CPCTC)

7) \overline{CA} bisects \angle ACD (if a segment splits an angle into two congruent parts, it is an angle bisector)

<u>Question 7</u>

1) \angle B and \angle C are right angles, \angle 4 \cong \angle 1 (given)

2) \triangle BAD and \triangle CAD are right triangles (definition of a right triangle)

3) \angle 1 \cong \angle 3 (vertical angles are congruent)

4) \angle 4 \cong \angle 3 (transitive property of congruence)

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

6) \therefore \triangle BAD \cong \triangle CAD (HA theorem)

7) \angle BDA \cong \angle CDA (CPCTC)

8) \therefore \vec{DA} bisects \angle BDC (definition of bisector of an angle)

8 0
2 years ago
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