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3 years ago

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Mathematics

1 answer:

Arturiano [62]3 years ago

5 0

Answer:

3/4 $\pi$ x 180/ $\pi$ = <u>135°</u>

Step-by-step explanation:

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Which statement is true

slava [35]

C.) because red is g and it goes up 6 right 6 an since blue is F it goes up 3 right 3 which would collide at the center

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2 years ago

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Marianne has 5/8 package of peas. She cooks 2/3 of those peas for 5 people. If each person is served an equal amount, what fract

creativ13 [48]

5/8 and 2/3 so...
8 times 3 is 24 so now put your fraction over twenty four
15/24=5/8
15/24 so fifteen out of twenty fourths of the package is what she has. She takes ten out of the fifteenths because fifteen divided by three is five then multiply by two is ten. Then divide by five and you get two.
15/24. 10-> 15/24 cause... 15 divided by 3 ( from the two thirds she cooked) is 5 then * by 2... 5*2=10

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2 years ago

If you roll a 6-sided die 4 times, what is the probability of one roll being a multiple of 3? A)2/3 B)16/81 C)145/1296 D)65/81 E

DerKrebs [107]

Answer:

671/1296

The probability of rolling at least one six is therefore 1 − 625/1296 = 671/1296 ≈ . 517.

Step-by-step explanation:

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3 years ago

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Solve for c 6/15=2/c

saul85 [17]

$\frac{6}{15}$ = $\frac{2}{c}$ Multiply both sides by 15
6 = $\frac{30}{c}$ Multiply both sides by c
6c = 30 Divide both sides by 6
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2 years ago

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