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tigry1 [53]
2 years ago
7

The triangles are similar, find the length of the unknown side. Round your answers to the nearest tenth (0.1), if necessary.

Mathematics
1 answer:
Doss [256]2 years ago
7 0

Answer:

22

Step-by-step explanation:

Similar shapes must have corresponding sides in a constant proportion. Therefore, we can set up the following equation and solve for ?:

\frac{28}{42}=\frac{?}{33} (divide corresponding sides)

\frac{28}{42}=\frac{?}{33},\\\\?=\frac{28\cdot 33}{42}=\boxed{22}

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3. In an experiment a scientist mixes together three substances. She mixes 18.42 g of
Harman [31]

Answer:

6.24

Step-by-step explanation:

A= 18.42g, B = 5.8g,C = 0.75g

Total = 18.42 + 5.8 + 0.75

= 24.97g

Then divided into four equal parts

= 24.97g/4

= 6.2425g

= 6.24g

7 0
3 years ago
PLS HELP!<br> solve by completing the square<br> x²+4x=8
masya89 [10]

Step-by-step explanation:

x²+4x=8

rearrange

x² + 4x - 8 = 0

half of 4 is 2

so

x² + 4x + 2² - 2² - 8 = 0

do x² + 4x + 2²

multiplies to give 4 adds to give 4

2+ 2

(x + 2)²

(x + 2)² - 2² - 8

(x + 2)² - 4 - 8

(x + 2)² - 12

(x + 2 + √12) (x + 2 - √12)

√12 = √4 * √3 = 2 √3

(x + 2 + 2√3) (x + 2 - 2√3)

4 0
2 years ago
5 positive integers are arranged in ascending order, as follows:
Helga [31]
The answer is: 10.

Explanation: The numbers are in order by the median/lowest to highest so in that case X wouldn't be 7 because the X would have been next to the 7 not on the other side of the 10.
4 0
2 years ago
Write an equation of the line passing through the points (−5, 6), (−7, 3)
blsea [12.9K]

Answer:

all work is shown and pictured

7 0
4 years ago
Can someone help me on this pls? It’s urgent, so ASAP (it’s geometry)
GarryVolchara [31]

<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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