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zubka84 [21]
2 years ago
10

Prove that the following triangles are similar?

Mathematics
1 answer:
Neporo4naja [7]2 years ago
7 0

Answer:

You have that:

angle N = angle Z

and

angle ZYX = angle NYM (because they are vertical angles

Since you have two congruent angles, you can say that the two triangles are similar according to the AA Triangle Similarity Postulate.

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20×(5×(-16) ) = (20×5)×(-16)​
stich3 [128]

Answer:

20×(5×(-16) = (20×5)×(-16)​

Step-by-step explanation:

They both equal -1600

7 0
3 years ago
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A withdrawal of $50 is negative or positive?
prisoha [69]
A withdrawal of $50 is negative because you are taking money out of your account and your money in your account is getting less than what it was before.
8 0
3 years ago
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1. Exercise 18, section 5.4. Compute 90, 91, 92, 93, 94 and 95. Make a conjecture about the units digit of 9n where n is a posit
Burka [1]

Answer:

n is a positive integer and here 90, 91, 92, 93, 94 and 95 is 9⁰ = 1

9¹ = 9

9² = 81

9³ = 729

9⁴ = 6561

9⁵ = 59049

so mathematical induction is prove given below.

Step-by-step explanation:

9⁰ = 1

9¹ = 9

9² = 81

9³ = 729

9⁴ = 6561

9⁵ = 59049

lets us consider by P (n) for the base case n = 0 then 9⁰ = 1  So P (1) is true

k ≥ 1so consider all integers

1≤l≤ k

There is a need to prove P (k+1)

if k is odd

9^k^+^1 =9^k.9.

9^k is 9 so

9^k = 9 +10m

then

9^k^+^1 = 9 ( 9 + 10m)

       =81 + 10 (9m)

so digit is 1 and k +1 is even

3 0
3 years ago
X^2 -3x + 0 = 0 Factor and solve
Luda [366]
So if you subtract the 0 out from the left side, the equation is now x^{2}-3x=0. Since a x is common in both of the terms on the left side, you can factor it out, now making the equation look like x(x-3)=0. When factoring, each of the separate things (x and (x-3)), must equal zero while the other one doesn't matter. So in order for the equation to be 0, the only potential options for x is 0 and 3. Therefore x=0 and x=3 are the two solutions
3 0
3 years ago
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Hello, I am currently very stuck with this problem and I am unsure as to how I would solve it.
borishaifa [10]

We have the equation

20y=x^2-10-15

Let's complete the square, to do it let's add and subtract 25 on the right side

\begin{gathered} 20y=x^2-10-15+25-25 \\  \\ 20y=(x-5)^2-15-25_{} \\  \\ 20y=(x-5)^2-40 \\  \\  \end{gathered}

Now we can have y in function of x

\begin{gathered} y=\frac{1}{20}(x-5)^2-2 \\  \\  \end{gathered}

Now we can already identify the vertex because it's in the vertex form:

y=a(x-h)+k

Where the vertex is

(h,k)

As we can see, h = 5 and k = -2, then the vertex is

(5,-2)

Now we can continue and find the focus, the focus is

\mleft(h,k+\frac{1}{4a}\mright)

We have a = 1/20, therefore

\begin{gathered} \mleft(5,-2+5\mright) \\  \\ (5,3) \end{gathered}

The focus is

(5,3)

And the last one, the directrix, it's

y=k-\frac{1}{4a}

Then

\begin{gathered} y=-2-5 \\  \\ y=-7 \end{gathered}

Hence the correct answer is: vertex (5, -2); focus (5, 3); directrix y = -7

5 0
1 year ago
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