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3 years ago

Use the image to answer the question.

Mathematics

1 answer:

Nonamiya [84]3 years ago

7 0

Insert a picture on the question bc there is no image

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Suppose a kite has two congruent angles of 100°. Of the remaining angles, one is four times larger than the other. What is the m

Pani-rosa [81]

Answer:

The measure of the largest angle in the kite is 128 degrees

Step-by-step explanation:

we know that

The sum of the interior angles in any quadrilateral must be equal to 360 degrees

A kite is a quadrilateral

Let

A,B,C and D the measure of the interior angles

$A+B+C+D=360^o$

we have that

$A=B=100^o$

substitute

$100^o+100^o+C+D=360^o$

$C+D=160^o$ ----> equation A

Remember that

one is four times larger than the other

$C=4D$ ----> equation B

substitute equation B in equation A

$4D+D=160$

solve for D

$5D=160\\D=32^o$

Find the measure of angle C

$C=4(32)=128^o$

therefore

The measure of the largest angle in the kite is 128 degrees

5 0

3 years ago

What's the temperature change?

Alex787 [66]

A. +40

B. -19

C. +130

D. +53

5 0

3 years ago

A square base that covers an area of 25 ft squared what is the length of one side of the tent?

sukhopar [10]

I was also going to say 6.25 but someone else did :)

5 0

3 years ago

Irina-Kira [14]

1. 5x/2 +y=-3
2. False
3. y= -5x-37 because y=mx-b
4. y= 1/5 x - 1 because y=mx-b
M= the slope and b represents the intercept.

5 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

3 years ago