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3 years ago

Determine the relationship between the two triangles and whether or not they can be proven to be congruent.

Mathematics

1 answer:

Step2247 [10]3 years ago

8 0

I think they are congruent because two of their lines are the same and the both have a 90 degree angle. I think they are congruent from the SAS postulate.

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Simplifying Algebraic Expressions ; 7+9(-6+8)

Grace [21]

Start by multiplying the 9 out:

9(-6 + 8y) = -54 + 72y

Then, add the 7:

-54 + 72y + 7

Then simplify the whole thing:

-47 + 72y

7 0

3 years ago

Help find the answer to what B is.

satela [25.4K]

Assuming that this is a right triangle, then (50 in)^2 = b^2 + (14 in)^2, according to the Pythagorean Theorem.

Then 196 in^2 + b^2 = 2500 in^2. Solving for b^2:

b^2 = (2500-196) in^2, and so b = +48 inches (answer)

7 0

3 years ago

you and four of your nine classmates received an A on your math test, what is the simplest fraction of students that scored an A

alex41 [277]

The simplest fraction is 4/5

8 0

3 years ago

Read 2 more answers

The perimeter of a rectangle is 152m. If the length is one morethan twice of its width. Find the length and width of rectangle.

sergejj [24]

Step-by-step explanation:

Perimeter of a Rect.= 2(l+b)

b= x

l= 2x

Equation

152= 2(2x+x)

152= 2(3x)

152= 6x

x= 25.33

2x= 50.66

Therefore length is 50.66m and breadth is 25.33m

3 0

3 years ago

A computer chess game and a human chess champion are evenly matched. They play twelve games. Find probabilities for the followin

sleet_krkn [62]

Answer:

The probability that they each win six games is 0.225.

Step-by-step explanation:

Given : A computer chess game and a human chess champion are evenly matched. They play twelve games.

To find : The probability that they each win six games?

Solution :

Applying binomial distribution,

Here n=12 and p=0.5

$P(X=k)=\frac{n!}{k!(n-k)!}\times p^k\times (1-p)^{n-k}$

The probability that they each win six games is k=6.

$P(X=6)=\frac{12!}{6!(12-6)!}\times 0.5^6\times (1-0.5)^{12-6}$

$P(X=6)=\frac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{6\times 5\times 4\times 3\times 2\times 6!}\times 0.015625\times 0.015625$

$P(X=6)=11\times 2\times 3\times 2\times 7\times 0.015625\times 0.015625$

$P(X=6)=0.225$

Therefore, The probability that they each win six games is 0.225.

3 0

3 years ago