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3 years ago

What is the perimeter of the triangle

Mathematics

1 answer:

podryga [215]3 years ago

5 0

Answer: it’s a really tiny picture so I hope I’m right (sorry if I’m not) 42

Step-by-step explanation:

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2 Points<br> Evaluate 5 – (4 + 2)2

Ghella [55]

Answer:

-31

Step-by-step explanation:

5 - (4 + 2)^2 = -31

6 0

4 years ago

Sarah has three skirts black white and blue and three blouses red green and yellow how many different outfits can she make if ea

zimovet [89]

Well if you think about it its 9 because

3 Skirts
3 Blouse

multiply it to get you answer and

3x3=9

8 0

3 years ago

Someone help plz. quick. :/

Katyanochek1 [597]

Answer:

y = 2x - 22

Step-by-step explanation:

y = 2x - 8

it is the form of y = mx + c

to be parallel the m should be the same ..

so the equation m = 2

znd it is given that it is passing (11,0) so that the equation should satisfy the point ( 11,0)

y = mx + c

y = 0 , m = 2 , x = 11

0 = 2 ( 11 ) + c

c = -22

so the required equation for the parallel line is

y = 2x - 22

6 0

3 years ago

Find the product of ( 2a2– 3b2) and ( a +b2- 3z)

Bond [772]

Answer:

$2a^3+2a^2b^2-6a^2z-3ab^2-3b^4+9b^2z$

Step-by-step explanation:

$(2a^2-3b^2)(a+b^2-3z)$

Start by distributing the $2a^2$ into all of the terms in between the second parentheses.

$2a^3+2a^2b^2-6a^2z$

Next, distribute the $-3b^2$ into all of the terms in between the second parentheses.

$-3ab^2-3b^4+9b^2z$

For the sake of simpler interpretation, I separated the two above, but they are together:

$2a^3+2a^2b^2-6a^2z-3ab^2-3b^4+9b^2z$

Since the terms are already in descending order, we do not need to rearrange them.

4 0

3 years ago

15 players for a softball team show up for a game: (a) How many ways are there to choose 10 players to take the field? 3003 (b)

prohojiy [21]

Answer:

(a) 3003 ways

(b) 10897286400 ways

Step-by-step explanation:

Given

$n = 15$ --- 15 players

Solving (a) Ways of selecting 10 players.

This implies combination.

So, we have:

$r=10$

Using:

$^nC_r = \frac{n!}{(n-r)!r!}$

We have:

$^{15}C_{10} = \frac{15!}{(15-10)!10!}$

$^{15}C_{10} = \frac{15!}{5!10!}$

Simplify

$^{15}C_{10} = \frac{15*14*13*12*11*10!}{5!10!}$

$^{15}C_{10} = \frac{15*14*13*12*11}{5!}$

$^{15}C_{10} = \frac{15*14*13*12*11}{5*4*3*2*1}$

$^{15}C_{10} = \frac{360360}{120}$

$^{15}C_{10} = 3003$

Solving (b) Ways of assigning positions to 10 players.

This implies permutation.

So, we have:

$r=10$

Using:

$^nP_r = \frac{n!}{(n-r)!}$

We have:

$^{15}P_{10} = \frac{15!}{(15-10)!}$

$^{15}P_{10} = \frac{15!}{5!}$

Solve each factorial

$^{15}P_{10} = \frac{1307674368000}{120}$

$^{15}P_{10} = 10897286400$

Solving (c) Ways of choosing at least 1 woman

We have:

$Men = 10$

$Women = 5$

Ways of selecting 10 players is: (a) 3003 ways

Since the number of men are 10, there is 1 way of selecting 10 men (i.e. selection without women)

Using complement rule:

At least 1 woman = Total - No woman

$At\ least\ 1\ woman = 3003 - 1$

$At\ least\ 1\ woman = 3002$

5 0

3 years ago

What is the perimeter of the triangle​

What is the perimeter of the triangle