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MrMuchimi
3 years ago
7

Please help!! I'll give brainliest! Identifying Triangle Congruence​

Mathematics
2 answers:
son4ous [18]3 years ago
8 0

Answer: your answers would be

1, SAS

2, SS

3,SAS

4, SAA

5, SAS

6, SA

7, SAS

8, SSS

9, SSA

10, ASA

11, SSS

12. ASA                                                                                                                    Explanation: A shape is congruent when it's all sides are the same SSS but also the shape's side and angle could also be counted as congruent. it can  be like this ASA, SAS, HL, and AAS are a theorem of Congruent but not AA, SS, SSA, AS and AAA this is because They have only two side congruent we don't know the 3rd side. Also SSA is not congruent because we don't know if the unknown side is equal another thing is SSA spelled back word is A## so lets just leave it at that

ddd [48]3 years ago
7 0

Answer:

1. SAS

2. SSS

3. ASA

4. AAS

5. RHS

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Please help on this one ?
Nastasia [14]

Answer:

C

Step-by-step explanation:

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it is a graph of the function given.

5 0
3 years ago
Which of the following is not an example of a parent function
marta [7]
A parent function is normally not streched or anything or shhifted

y=kx or
y=x^a or
y=a^x are exapmles

y=x is y=1x
y=2x^3 is not it because the parent would be y=x^3
y=1/x is y=x^-1
y=2^x is y=2^x



answer is 2x^3 is not a parent function
5 0
3 years ago
Brandy has a collection of comic books. If she adds 15 to the number of comic books in her collection and
maxonik [38]
Let the collection of comic books be x.
4 times the number of comic books in her collection = 4x

3( x + 15 ) = 4x - 65
3x + 45 = 4x - 65
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ans : 110
4 0
3 years ago
Write the equation of the line, with the given properties, in slope-intercept form.
Vinvika [58]

Answer:

y=-7x-30

Step-by-step explanation:

y-y1=m(x-x1)

y-5=-7(x-(-5))

y-5=-7(x+5)

y=-7x-35+5

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3 0
3 years ago
A.Find a formula for
snow_lady [41]

Answer:

a) \frac{n}{n+1}

b) Proof in explanation.

Step-by-step explanation:

a)

\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)}.

So let's look at the last term for a minute:

\frac{1}{n(n+1)}

Let's use partial fractions to see if we can find a way to write this so it is more useful to us.

\frac{1}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}

Multiply both sides by n(n+1):

1=A(n+1)+Bn

Distribute:

1=An+A+Bn

Reorder:

1=An+Bn+A

Factor:

1=n(A+B)+A

This implies A=1 and A+B=0 which further implies that B=-1.

This means we are saying that:

\frac{1}{n(n+1)} can be written as \frac{1}{n}+\frac{-1}{n+1}

We can check by combing the fractions:

\frac{n+1}{n(n+1)}+\frac{-n}{n(n+1)}

\frac{n+1-n}{n(n+1)}

\frac{1}{n(n+1)}

So it does check out.

So let's rewrite our whole expression given to us using this:

(\frac{1}{1}+\frac{-1}{2})+(\frac{1}{2}+\frac{-1}{3})+(\frac{1}{3}+\frac{-1}{4})+\cdots +(\frac{1}{n}+\frac{-1}{n+1})

We should see that all the terms in between the first and last are being zeroed out.

That is, this sum is equal to:

\frac{1}{1}+\frac{-1}{n+1}

Multiply first fraction by (n+1)/(n+1) so we can combine the fractions:

\frac{n+1}{n+1}+\frac{-1}{n+1}

Combine fractions:

\frac{n}{n+1}

b)

Proof:

Let's see what happens when n=1.

Original expression gives us \frac{1}{1 \cdot 2}=\frac{1}{2}.

The expression we came up with gives us \frac{1}{1+1}=\frac{1}{2}.

So it is true for the base case.

Let's assume our expression and the expression given is true for some integer k greater than 1.

We want to now show it is true for integer k+1.

So under our assumption we have:

\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \frac{1}{k(k+1)}=\frac{k}{k+1}

So let's add the (k+1)th term of the given series on both sides:

\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}

(Now we are just playing with right hand side to see if we can put it in the form our solution which be if we can \frac{k+1}{k+2}.)

I'm going to find a common denominator which will be (k+1)(k+2):

\frac{k}{k+1} \cdot \frac{k+2}{k+2}+\frac{1}{(k+1)(k+2)}

Combine the fractions:

\frac{k(k+2)+1}{(k+1)(k+2)}

Distribute:

\frac{k^2+2k+1}{(k+1)(k+2)}

Factor the numerator:

\frac{(k+1)^2}{(k+1)(k+2)}

Cancel a common factor of (k+1)

\frac{k+1}{k+2}

We have proven the given expression and our formula for the sum are equal for all natural numbers,n.

6 0
2 years ago
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