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borishaifa [10]
2 years ago
11

Find the value of x and then classify the triangles

Mathematics
1 answer:
ZanzabumX [31]2 years ago
8 0
6. x = 16; it's a scalene triangle

7. x = 45; it's a right triangle. 
You might be interested in
Problem of the Day:
Shtirlitz [24]

Answer:

Step-by-step explanation:

What this question is asking of you is what is the greatest common divisor of 12 and 15. Or, what is the biggest number that divides both 12 and 15.

in order to find this we have to split each number into it's prime components.

for 12 they are 2,2 and 3 (

2

⋅

2

⋅

3

=

12

)

and for 15 they are 3 and 5 (

3

⋅

5

=

15

)

Out of those two groups (2,2,3) and (3,5) the only thing in common is 3, so 3 is the greatest common divisor. That tells us that the greatest number of groups that can exist and have the same number of girls and the same number of boys for each group is 3.

Now to find out how many girls and boys there are going to be in each group we divide the totals by 3, so:

12

3

=

4

girls per group, and

15

3

=

5

boys per group.

(just as a thought exercise, if there were 16 boys, the divisors would have been (2,2,3) and (2,2,2,2), leaving us with 4 groups [

2

⋅

2

] of 3 girls [12/4] and 4 boys [16/4] )

8 0
3 years ago
Which set of numbers could be the lengths of the sides of a triangle?
tester [92]

The correct answer would be A), because the two shortest sides should be greater than the longest side when added together.

4 0
3 years ago
A/20 + 14/15 = 9/15 <br><br> Can i get some help?
KiRa [710]
Method 1:

\dfrac{a}{20}+\dfrac{14}{15}=\dfrac{9}{15}\ \ \ \ |multiply\ both\ sides\ by\ 60\\\\60\cdot\dfrac{a}{20}+60\cdot\dfrac{14}{15}=60\cdot\dfrac{9}{15}\\\\3a+4\cdot14=4\cdot9\\\\3a+56=36\ \ \ \ |subtract\ 56\ from\ both\ sides\\\\3a=-20\ \ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{a=-\frac{20}{3}\to a=-6\frac{2}{3}}

Method 2.

\dfrac{a}{20}+\dfrac{14}{15}=\dfrac{9}{15}\ \ \ \ |subtract\ \dfrac{14}{15}\ form\ both\ sides\\\\\dfrac{a}{20}=-\dfrac{5}{15}\\\\\dfrac{a}{20}=-\dfrac{1}{3}\ \ \ \ |multiply\ both\ sides\ by\ 20\\\\\boxed{a=-\dfrac{20}{3}\to a=-6\frac{2}{3}}
3 0
3 years ago
Q3. A bag contains 7 red balls and 3 green balls. A ball is taken out replaced.
ycow [4]

Part (a)

There are 7 red out of 7+3 = 10 total

<h3>Answer: 7/10</h3>

==========================================================

Part (b)

We have 3 green out of 10 total

<h3>Answer: 3/10</h3>

==========================================================

Part (c)

3/10 is the probability of getting green on any selection. This is because we put the first selection back (or it is replaced with an identical copy)

So (3/10)*(3/10) = 9/100 is the probability of getting two green in a row.

<h3>Answer: 9/100</h3>

==========================================================

Part (d)

Similar to part (c) we have 7/10 as the probability of getting red on each independent selection.

(7/10)*(7/10) = 49/100

<h3>Answer: 49/100</h3>

==========================================================

Part (e)

7/10 is the probability of getting red and 3/10 is the probability of getting green. Each selection is independent of any others.

(7/10)*(3/10) = 21/100

<h3>Answer: 21/100</h3>

==========================================================

Part (f)

We have the exact same set up as part (e). Notice how (7/10)*(3/10) is the same as (3/10)*(7/10).

<h3>Answer: 21/100</h3>
7 0
2 years ago
On a piece of paper graph......
Artyom0805 [142]
Answer:

I think it’s the first one, A
6 0
3 years ago
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