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2 years ago

How does labeling the diagram help us show that we can use right triangle congruency (SAS in this case)?

Mathematics

1 answer:

dem82 [27]2 years ago

4 0

Answer:

Step-by-step explanation:

S A S congruence rule: Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of other triangle.

If two triangles are congruent by SAS, and the angle is 90, then by CPCT{corresponding part of congruent triangle) the hypotenuse of the two triangles will be congruent.

Now, as hypotenuse and one leg are congruent in both triangles, we can apply RHS congruent

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3 years ago

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Fudgin [204]

Answer: Function, not a function, Function, not a function

Step-by-step explanation:

For it to be a function, each value of x has to map onto one value of y.

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1 year ago

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2 years ago

Please please help. I have a bunch of these and I don’t understand it at all

shtirl [24]

The correct option is: Option (B) $\sum_{i=1}^{7} (5i-2)$

Explanation:

First thing is that the difference between each number in the series with the next number is 5. It means it must be the multiple of 5. There are two options that contain multiples of 5: Option B and Option D. Now in the option D, the upper limit is 6. If we put 6 in the expression: 5(6)-2, the last term would be 28. However in the series given in the question, the last term is 33. Hence 5(7) - 2 = 35 - 2 = 33 which is Option B.

When i=1: 5(1)-2 = 3
When i=2: 5(2)-2 = 8
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.
.
When i=7: 5(7)-2 = 35-2 = 33
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3 years ago

The nth term of a sequence is 3n^2- 1

Thepotemich [5.8K]

Answer: The number that belongs to both sequences is 26.

Step-by-step explanation:

We have two sequences, let's call one as A and the other as B.

The n-th term of sequence A is written as:

aₙ = 3*n^2 - 1

the nth term of sequence B is written as:

bₙ = 30 - n^2

We want to find a term that belongs to both sequences, (it can be for different integers, we can use n for sequence A and x for sequence B)

Then we want to find:

aₙ = bₓ

where n and x are integer numbers.

Then we will heave:

3*n^2 - 1 = 30 - x^2

To find the pair, we could isolate one of the variables, then input different integers in the other variable and see if the outcome is also an integer.

Let's isolate n.

3*n^2 = 30 - x^2 + 1

3*n^2 = 31 - x^2

n^2 = (31 - x^2)/3

n = √( (31 - x^2)/3)

Now let's input different values for x, and see if the outcome is also an integer, notice that x is in a negative term inside a square root, then we have only a few values of x such that the equation can be true.

Then let's start with x = 1.

n(1) = √( (31 - 1^2)/3) = √(30/3) = √10

We know that √10 is not an integer.

now with x = 2,

n(2) = √( (31 - 2^2)/3) = √( (31 - 4)/3) = √(27/3) = √9 = 3

then if x = 2, we have n = 3.

Both of them are integers, then we get:

a₂ = 3*(3)^2 - 1 = 27 - 1 = 26

b₃ = 30 - 2^2 = 30 - 4 = 26

The number that belongs to both sequences is 26.

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2 years ago