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3 years ago

Given AB intersects DE at point C. prove: DCB = ECA. What is the missing reason in step 5

Mathematics

1 answer:

kicyunya [14]3 years ago

5 0

Answer: the answer is linear pair

Step-by-step explanation:

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the rational roots of a polynomial function f(x) can be written in the form p/q where p is a factor of the leading corfficient o

Lorico [155]

False.

The Rational Root theorem states that P is a factor of the constant term and q is a factor of the leading coefficient.

4 0

3 years ago

Kris and his dad built an ice rink. The area is 74.4 sq meters. The length is 12 meters what is the width

-BARSIC- [3]

Answer:

6.2 meters

Step-by-step explanation:

1) area = l*w

74.4 = 12x

x=6.2

3 0

2 years ago

Cyril put a total of 1/8 lb of gravel into 6 fish tanks. He out the same amount cof gravel into each tank. How many pounds if gr

8090 [49]

(1/8) / 6 =
1/8 * 1/6 =
1/48 of a lb per fish tank <==

4 0

3 years ago

I Am Not Sure if B is Correct Please Help Me

Nookie1986 [14]

YZ is the diameter, so arc YZ = 360°/2 = 180°

arc XZ = 360° - arc YZ - arc YWX = 360° - 180° - 100° = 80°

∠XYZ is an inscribed angle <span>so it is equal to half the arc XZ

</span>∠XYZ = 80°/2 = 40°

3 0

3 years ago

How do you do this question?

alina1380 [7]

Answer:

(8√2) / 15

Step-by-step explanation:

A curve bounded by the y-axis is represented by in terms of dy;

$\int \:x\:dt$

When the curve crosses the y-axis, x will be 0. In this case x is the function of t, so we have to solve for x(t) = 0;

0 = t^2 + 2t --- (1)

Solution(s) => t = 0, t = 2

dy = (1/2 * 1/√t)dt --- (2)

Our solutions (0, 2) are our limits. The area of the curve is in the form $A\:=\:\int _b^a\:f\left(t\right)g'\left(t\right)dt$ , so now let's introduce the limits of integration, x(t) and dy/dt. Remember, dy/dt = (1/2 * 1/√t) (second equation). 1/2 * 1/√t can be rewritten as 1/2 * t^(-1/2)....

$A\:=\:\int _2^0\:\left(t^2-2t\right)\left(\frac{1}{2}t^{-\frac{1}{2}}\right)dt\\\\= \int _2^0\:\left(\frac{1}{2}t^{\frac{3}{2}}-t^{\frac{1}{2}}\right)dt\\\\= \left[\frac{t^{\frac{5}{2}}}{5}-\frac{2t^{\frac{3}{2}}}{3}\right]_2^0\\\\= 0\:-\:\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\\\\= \frac{8\sqrt{2}}{15}$

Your solution is 8√2 / 15

7 0

3 years ago