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3 years ago

Geometry math question no Guessing and Please show work

Mathematics

2 answers:

Jlenok [28]3 years ago

8 0

D because the sum of 2 sides of a triangle have to be greater than the 3rd side so 7+4 is the only option where the sum of both sides will be greater than the 3rd side

Pavlova-9 [17]3 years ago

6 0

This question isn't very specific. Are we assuming that this is a right triangle and calculating the hypotenuse.

I'm going to guess not, because the hypotenuse is clearly the longest length, and all the choices are not above 10.

So, what this means is we are calculating B.

We make the equation:

7 $7^2+b^2=10^2$

From here, we solve and get: $49 + b^2 = 100 \\ 100-49 = b^2 \\ 51 = b^2 \\ \sqrt{51} \\ b = 7.14$

So, clearly this is not an option.

Is there any additional information provided? Is it a special triangle?

This would be right if it was a right triangle. Let me know if I wrong in the comments.

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Write the ways to solve it too plz

xenn [34]

A.) -540

Sum = N (first number + last number) / 2

N = count of consecutive numbers. Had the numbers included 1, then, N would have been 33. Since its range is 7 - 33 then 33 - 7 + 1= 27.
N = 27

Sum = 27 (-33 - 7) / 2
sum = 27 (-40)/2
sum = -1080 / 2
sum = -540

3 0

3 years ago

A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 y

Anestetic [448]

Answer:

$\frac{1}{4180}$

Step-by-step explanation:

GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are $12$ red candies, $12$ blue, $7$ green, $13$ brown, $3$ orange, and $10$ yellow. Three candies are pulled from the bag in succession, without replacement.

TO FIND: What is the probability that the first two candies drawn are orange and the third is green.

SOLUTION:

Total candies in the bag $=57$

Probability that first ball is orange, $P(A)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}$

$=\frac{3}{57}=\frac{1}{19}$

Probability that second ball is orange, $P(B)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}$

$=\frac{2}{56}=\frac{1}{28}$

Probability that third ball is green, $P(C)=\frac{\text{total green candies in bag}}{\text{total candies in bag}}$

$=\frac{7}{55}$

Now, probability that first two balls are orange and third is green is

$=P(A)\times P(B)\times P(C)$

$=\frac{1}{19}\times\frac{1}{28}\times\frac{7}{55}$

$=\frac{1}{19}\times\frac{1}{4}\times\frac{1}{55}$

$=\frac{1}{4180}$

Hence, probability that first two balls are orange and third is green is $\frac{1}{4180}$

3 0

4 years ago

In a box of crayons, 24% of the crayons are blue.There are 48 blue crayons in the box.

-BARSIC- [3]

D. There are 200 crayons in the box. If you just made it 25% of 200 is 50 which is the closest number to answer based on a quick estimation:)

5 0

3 years ago

I need help on number 4

Svetach [21]

Ok
Part B: Both of the children are incorrect. The children have the same amount of money
Explanation: they have the same amount because 6 dimes are 60 cents... Hannah says she has 60 cents while Mason says he has 6 dimes. Got it?
Part A: 6 dimes = 60 cents
Explanation: None

6 0

4 years ago

What is 13000000000 in scientific notation

xxTIMURxx [149]

<span>notation of thirteen million.

hope this helps you(:</span>

5 0

3 years ago

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