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Whitepunk [10]
2 years ago
14

I need help with math can someone help me I’ll show the question?

Mathematics
1 answer:
Alecsey [184]2 years ago
5 0
What’s your question
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Consider quadrilateral BCEF inscribed in circle A. Diagonals EB and CF intersect at point D. Select all the statements that are
Elena L [17]

Answer:

m\angle ECB+m\angle EFB=180

m\angle CDB\cong m\angle EDF

m\angle CDB+m\angle DCB+m\angle CBD=180 \degree

Step-by-step explanation:

From the diagram, quadrilateral BCEF is a cyclic quadrilateral.

Opposite angles if a cyclic quadrilateral sum up to 180°

m\angle ECB+m\angle EFB=180

The diagonals intersect at D to form two pairs of vertical angles, and vertical angles are congruent.

m\angle CDB\cong m\angle EDF

Also sum of angles in triangle CBD is 180°.

m\angle CDB+m\angle DCB+m\angle CBD=180 \degree

6 0
2 years ago
Find the slope and y intercept
alexgriva [62]
Slope: 5/1 (5 if simplified)
y-intercept: 10
7 0
3 years ago
Which of the following is not a fee that contributes to the initial cost of leasing a car?
Harlamova29_29 [7]
Here are the answers to the questions above:
1. Based on the list of fees above, the one that does not contribute to the <span>initial cost of leasing a car is the FINAL PAYMENT. Answer would be option B.
2. In leasing a car, the amount that does not affect the total cost is the PRINCIPAL CHARGE, and the answer for this is option A. 
Hope this helps.</span>
3 0
3 years ago
Read 2 more answers
What is the ordered pair solution for this system of equation X+y=8<br> Y=x-3
Korvikt [17]

Answer:

(5.5, 2.5)

Step-by-step explanation:

Plug in X and Y and guess is how I did it

6 0
3 years ago
The pressure applied to a leverage bar varies inversely as the distance from the object. If 150 pounds is required for a distanc
AlexFokin [52]

Answer:

500 pounds

Step-by-step explanation:

Let the pressure applied to the leverage bar be represented by p

Let the distance from the object be represented by d.

The pressure applied to a leverage bar varies inversely as the distance from the object.

Written mathematically, we have:

p \propto \dfrac{1}{d}

Introducing the constant of proportionality

p = \dfrac{k}{d}

If 150 pounds is required for a distance of 10 inches from the object

  • p=150 pounds
  • d=10 inches

150 = \dfrac{k}{10}\\\\k=1500

Therefore, the relationship between p and d is:

p = \dfrac{1500}{d}

When d=3 Inches

p = \dfrac{1500}{3}\\\implies p=500$ pounds

The pressure applied when the distance is 3 inches is 500 pounds.

4 0
3 years ago
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