Given the following: measure of arc A B equals measure of arc B C equals measure of arc C D.

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

5 0

In circle F, what is the measure of ∠ CED ?
Given mAB = mBC = mCD = 36°

So
 ∠ CED = 1/2(mCD) = 1/2(36°) = 18°

Answer:
B. 18°
---------------------------------------------------------------------

What is the measure of ∠GET?

Given mTG = 70°
so
<TNE = 1/2(mTG) = 1/2(70°) = 35°

Right triangle NTG , < NTE = 90° and <TNE = 35°
so <TEN = 90° - 35° = 55°

<TEN = <GET = 55°

Answer
A. 55°

You might be interested in

Use the divergence theorem to find the outward flux of the vector field F(x,y,z)=2x2i+5y2j+3z2k across the boundary of the recta

aksik [14]

Answer:

The answer is "120".

Step-by-step explanation:

Given values:

$F(x,y,z)=2x^2i+5y^2j+3z^2k \\$

differentiate the above value:

$div F =2 \frac{x^2i}{\partial x}+5 \frac{y^2j}{\partial y}+3 \frac{z^2k }{\partial z} \\$

$= 4x+10y+6z$

$\ flu \ of \ x = \int \int div F dx$

$= \int\limits^1_0 \int\limits^3_0 \int\limits^1_0 {(4x+10y+6z)} \, dx \, dy \, dz \\\\ = \int\limits^1_0 \int\limits^3_0 {(4xz+10yz+6z^2)}^{1}_{0} \, dx \, dy \\\\ = \int\limits^1_0 \int\limits^3_0 {(4x+10y+6)} \, dx \, dy \\\\ = \int\limits^1_0 {(4xy+10y^2+6y)}^3_{0} \, dx \\\\ = \int\limits^1_0 {(12x+90+18)}\, dx \\\\= {(12x^2+90x+18x)}^{1}_{0} \\\\= {(12+90+18)} \\\\=30+90\\\\= 120$