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muminat
2 years ago
13

A rectangular prism has a volume is 98 cubic feet a, a width of 2 feet and a length of 7 feet. Find the height of the regular pr

ism
Mathematics
1 answer:
mariarad [96]2 years ago
5 0

In the construction world, most materials are measured in cubic yards. Multiply the length (L), in feet, by the width (W), in feet, by the height (H), in feet, and divide by 27. This number is how many cubic yards of crushed stone you need.

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Find all positive integer values of c that make x^2+6x+c factorable
densk [106]
It's factorable when c  ∈ {5,8,9}
6 0
3 years ago
Can someone please help me on this one
likoan [24]

You know vertical Angles SRU and TRV are congruent.

You are given that Sides UR and VR are congruent.

You are given that Angles SUT and SVT are congruent.

An appropriate choice is the ASA postulate, since you have congruent angles with congruent sides in between.

4 0
3 years ago
The length of a rectangle is 10 meters more than the width. If 3 meters is added to both the length and the width, the new area
Finger [1]

Answer:

L = 20

W = 10

Step-by-step explanation:

L = (w + 10)

NL = (w + 13)

Divide 299 by 13:

299 ÷ 13 = 23

So, 23 x 13 = 299

L = 23

W = 13

Subtract 3 from both to get the original length and width

4 0
3 years ago
If ∠C and ∠D are complementary angles, m∠C = (4x + 3)°, find m∠D = (15x - 8)°, find m∠D.
kykrilka [37]

Answer:

67

Step-by-step explanation:

4x+3+15x-8=90(complementary angle)

x=5

now D = 15*5-8

d=67

8 0
3 years ago
*In ∆ABC, on the extension of the side BC , draw a line segment CD ≅ CA . Draw the segment AD . The line segment CE is the angle
Leya [2.2K]

CE ⊥ CF because m∠ECA + m∠FCA = 90°

Step-by-step explanation:

In ∆ABC

  • On the extension of the side BC , draw a line segment CD ≅ CA
  • Draw the segment AD
  • The line segment CE is the angle bisector of ∠ACB
  • The line segment CF is the median towards AD in ∆ ACD

We want to prove that CF ⊥ CE

Look to the attached figure

In Δ ABC

∵ CE is the bisector of angle ACB

∴ ∠ACE ≅ ∠BCE

In Δ ACD

∵ CA = CD

∴ Δ ACD is an isosceles triangle

∵ AD is the median towards AD

- In any isosceles triangle the median from a vertex to its opposite

  side bisects this vertex

∴ AD bisects ∠ACD

∴ ∠ACF ≅ ∠DCF

∵ BCD is a straight segment

∵ CE , CA , CF are drawn from point C

∴ m∠BCE + m∠ACE + m∠ACF + m∠DCF = 180°

∵ m∠ACE ≅ m∠BCE

∵ m∠ACF ≅ m∠DCF

- Replace m∠BCE by m∠ACE and m∠DCF by m∠ACF

∴ m∠ACE + m∠ACE + m∠ACF + m∠ACF = 180°

∴ 2 m∠ACE + 2 m∠ACF = 180°

- Divide all terms by 2

∴ m∠ACE + m∠ACF = 90°

∴ EC ⊥ CF

CE ⊥ CF because m∠ECA + m∠FCA = 90°

Learn more:

You can learn more about perpendicular lines in brainly.com/question/11223427

#LearnwithBrainly

7 0
3 years ago
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