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2 years ago

Solve for x (geometry) ILL GIVE BRAINLIEST

Mathematics

1 answer:

Minchanka [31]2 years ago

8 0

Answer:

Step-by-step explanation:

jesus this is the third time you put up this question thx for free points ig

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RSM HW PLEASE HELPPPPPPPPPPPPP ASAP

Maksim231197 [3]

Answer:

congruent SAS

Step-by-step explanation:

We know two sides of the triangles are congruent to each other

MD = MT

and MA = MU

We also know that <DMA = < TMU

Two sides and the included angle

We can use SAS to show that the triangles are congruent

7 0

2 years ago

Read 2 more answers

Select the correct answer.

Pani-rosa [81]

Answer:

Step-by-step explanation:

8 0

2 years ago

Alia has swimming class every fourth day and diving class every sixth day. If she had swimming class and diving class on 5 june,

dolphi86 [110]

Answer:

June 17

Step-by-step explanation:

the least common multiple of 4 and 6 is 12. Add 12 to June 5 and you get June 17

3 0

2 years ago

Select the denominator of 4/9

Assoli18 [71]

The denominator is the one on the bottom so 9 is the denominator the top is the numerator

5 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago