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3 years ago

∠A and ∠B are vertical angles. If m∠A=(7x-20)° and m∠B=(6x-1)°, find the measure of ∠A. (Just type answer as a number.)

Mathematics

1 answer:

WINSTONCH [101]3 years ago

6 0

Answer:

x=19 so ∠A=113

Step-by-step explanation:

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3 years ago

Find the volume of a cylinder. Round your answer to the nearest tenth. Do not add a comma.

natali 33 [55]

Answer:

Step-by-step explanation:

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3 years ago

If DF=78, DE=5x-9, and EF=2x+10, find DE.

Kazeer [188]

Given that E is a point between Point D and F, the numerical value of segment DE is 46.

<h3>What is the numerical value of DE?</h3>

Given the data in the question;

E is a point between point D and F.
Segment DF = 78
Segment DE = 5x - 9
Segment EF = 2x + 10
Numerical value of DE = ?

Since E is a point between point D and F.

Segment DF = Segment DE + Segment EF

78 = 5x - 9 + 2x + 10

78 = 7x + 1

7x = 78 - 1

7x = 77

x = 77/7

x = 11

Hence,

Segment DE = 5x - 9

Segment DE = 5(11) - 9

Segment DE = 55 - 9

Segment DE = 46

Given that E is a point between Point D and F, the numerical value of segment DE is 46.

Learn more about equations here: brainly.com/question/14686792

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2 years ago

katrice bought three watermelons for $10 each and two buckets of ice cream for $8 each. The store had a special for 1 free water

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4 years ago

Read 2 more answers

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

8 0

4 years ago