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2 years ago

14

Can someone please help me? I'm desperate lol

2 answers:

ddd [48]2 years ago

7 0

Answer:

y = 3

Step-by-step explanation:

Since the equation is parallel to the x-axis, it is a horizontal line, which means it has a slope of 0. Since anything times 0 is 0, that means x times 0 is 0, so we can take that out of the equation. That only leaves the y-intercept, which is 3. Substituting it into the equation, we get y = 3.

Hope this helps!

If you still don't understand or need help, send a comment or message.

Usimov [2.4K]2 years ago

6 0

Answer:

y would simply be 3

Step-by-step explanation:

i’m not sure if i can

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Answer:

x=7, angle a= 45 degrees, angle b= 65 degrees

Step-by-step explanation:

All the angles in a triangle must add up to 180. Subtract 70 from 180, add both angle a and angle b to get 16x-2=110, solve for x, x=7. Substitute x out for 7 in the equations then simplify to get angle a=45 and angle b=65

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Step-by-step explanation:

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2 years ago

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2 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.

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