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jonny [76]
2 years ago
14

What is the measure of each angle of an equilateral triangle?

Mathematics
2 answers:
nasty-shy [4]2 years ago
3 0

Answer:

C) 60 degrees

Step-by-step explanation:

An equilateral triangle has all equal angles. We know that all the angles of a triangle add up to 180 degrees. Therefore 45+45+45=180 giving you an equilateral triangle angle measure.

atroni [7]2 years ago
3 0
The answer is 60 degrees I think!
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If you graph all, your answer is a
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deff fn [24]
B (second choice) it just makes sense when u think about it
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Graph the line of the equation 8x/5 - 2y = -8 using its slope and y-intercept
satela [25.4K]

Answer:

y=\frac{4}{5}x+4

Step-by-step explanation:

You'll need to get this equation in slope-intercept form by solving for y. I do a little extra here to get it in the correct form, but I think it's pretty clear. Let me know if I need to clarify.

\\\\\frac{8x}{5}-2y=-8\\\\-2y=-\frac{8x}{5}-8\\\\2y=\frac{8x}{5}+8\\\\y=\frac{4}{5}x+4

Once it's in slope-intercept form, both the slope and the y-intercept are readily available so you can easily graph it. I graphed both of them in the attached image so you can see that they are the same line.

5 0
2 years ago
Jennifer has 84.5 yards of fabric to make curtains she makes 6 identical curtains and has 19.7 yards of fabric remaining how man
harkovskaia [24]

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Step-by-step explanation:

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6 0
3 years ago
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Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

3 0
3 years ago
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