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MatroZZZ [7]
3 years ago
15

HELP PLEASE ASAP IM ON THE LAST ONE

Mathematics
1 answer:
inna [77]3 years ago
7 0

Answer:

y=9; z=16

Step-by-step explanation:

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Quadrilateral A'B'C'D' is the image of quadrilateral ABC D under a dilation.
Semenov [28]

Answer:

Step-by-step explanation:

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3 years ago
Determine if the two triangles are congruent. If they are, state how you know.
earnstyle [38]

Answer:

A. SAS

Step-by-step explanation:

Since the marked angle is between the marked sides, and the corresponding sides are marked as congruent, the two triangles are congruent by SAS (side angle side).

7 0
3 years ago
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Which of the following is equivalent to x + 1/y ÷ g + 1/h
Firlakuza [10]

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say what now?

Step-by-step explanation:

7 0
2 years ago
Laura has already baked 9 pies, and she canbake 2 pies with each additional cup ofsugar she buys. With 14 additional cups ofsuga
qaws [65]

let x represent the number of cups of additional cups of sugar and y represents the total number of pies she can bake.

Given that;

"Laura has already baked 9 pies, and she can bake 2 pies with each additional cup of sugar she buys"

y=9+2x

With 14 additional cups of sugar;

x=14

substituting the value of x;

\begin{gathered} y=9+2(14) \\ y=9+28 \\ y=37 \end{gathered}

Therefore, the total number of pies Laura can Bake with 14 additional cups of sugar is;

undefined

3 0
1 year ago
Use induction to prove that 2? ?? for any integer n>0 . Indicate type of induction used.
Hoochie [10]

Answer with explanation:

The given statement is which we have to prove by the principal of Mathematical Induction

    2^{n}>n

1.→For, n=1

L H S =2

R H S=1

2>1

L H S> R H S

So,the Statement is true for , n=1.

2.⇒Let the statement is true for, n=k.

      2^{k}>k

                   ---------------------------------------(1)

3⇒Now, we will prove that the mathematical statement  is true for, n=k+1.

     \rightarrow 2^{k+1}>k+1\\\\L H S=\rightarrow 2^{k+1}=2^{k}\times 2\\\\\text{Using 1}\\\\2^{k}>k\\\\\text{Multiplying both sides by 2}\\\\2^{k+1}>2k\\\\As, 2 k=k+k,\text{Which will be always greater than }k+1.\\\\\rightarrow 2 k>k+1\\\\\rightarrow2^{k+1}>k+1

Hence it is true for, n=k+1.

So,we have proved the statement with the help of mathematical Induction, which is

      2^{k}>k

                 

   

3 0
3 years ago
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