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3 years ago

En la figura anterior, las rectas M y N son paralelas. ¿Cuántos grados mide el ángulo BAC ?

Mathematics

1 answer:

erik [133]3 years ago

5 0

180-(110+40)=30

use two parallel lines and intersected line rule...

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Latoya drove 210 miles using 9 gallons of gas. At this rate how many gallons of gas would she need to drive 413 miles?

Minchanka [31]

This is a problem that could be solved with proportions. Here's how to set it up.

210mi./9 gallons =413mi./x gallons

Then, you would cross multiply 9x413 then divide that my 210. The result would be the X gallons. Good Luck ♡

3 0

2 years ago

Chin researched the amount of money 150 students earned per month from jobs held during the summer. He created a table of six sa

dem82 [27]

the choices are below

the predicted mean of the population will be less than 200.

The predicted mean of the population will be less than 245.

The predicted mean of the population will be more than 275.

The predicted mean of the population will be more than 250.

b is the answer

8 0

2 years ago

Read 2 more answers

Solve using the zero product property x^2+11x+30=0 can you help please

skelet666 [1.2K]

I think it's x=-5, -6

4 0

3 years ago

Is three and five a factor pair of 15

Pie

Yes because 3*5 = 15

Factors multiply to get the product.

6 0

3 years ago

Read 2 more answers

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

Put $(x,y)=(1,1)$

$\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9$

This is true.

So, the point $(1,1)$ satisfies the equation $\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

So, point H is $(1,1)$ .

7 0

2 years ago