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Ray Of Light [21]
3 years ago
10

Find the solution of the system of equations. -X - 7y = -41 X-6y=-37

Mathematics
1 answer:
Lelu [443]3 years ago
3 0

Answer:

Hi how are you doing today Jasmine

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If x = 14 cm, y = 7 cm, and z = 8 cm, what is the surface area of the figure?
Nat2105 [25]

Formula for sufrace area of rectangular prisms: S= 2 (lw + lh + wh)

for the first prism, x=14, plug 14 into 2(x)--> 2(14)=28

7,8,28

S= 2 [7(8)+7(28)+8(28)]

S= 2 (28+196+224)

S= 2(448)

S= 896

for the second prism, x is just 14

7,8,14

S= 2 [7(8)+7(14)+8(14)]

S= 2 (28+98+112)

S= 2(238)

S= 476

Add the two sufrace areas we found and the answer will be

D) S= 1,372

8 0
2 years ago
Anyone know what 4x + 2y - 6z -x -3y + 2z ?
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4x+2y-6z-x-3y+2z= 3x-y-4z

For more help use math/way/./com

6 0
3 years ago
Read 2 more answers
Use the quadratic formula to find the solution to the quadratic equation in the box below.
nekit [7.7K]
The solution to the quadratic equation is option D
4 0
2 years ago
Simplify this expression.
amm1812

Answer: 0.88

Step-by-step explanation:

Okay -5.37 + 8.14 = 2.77 ( you can simply plug this into the calculator or you could switch the problem around so it’s 8.14 - 5.37, either way you’ll get 2.77)

And then you take 2.77 -1.89 = 0.88

So 0.88 is your answer

7 0
2 years ago
A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow
Mrrafil [7]

Answer:

\displaystyle \frac{54}{5405}.

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

\displaystyle \left(50\atop 5\right) = 2,118,760.

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

\displaystyle \left( 28\atop 2\right) = 378.

Number of ways to choose 3 green candies out of a batch of 8:

\displaystyle \left(8\atop 3\right)=56.

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168.

The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:

\displaystyle \frac{21,168}{2,118,760} = \frac{54}{5405}.

3 0
2 years ago
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