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2 years ago

6

Which graph represent me y= y=3√x-5 ( I can’t add all 4 of the graphs because it won’t let me attach 2 photos) BUT I REALLY NEED

HELP!!

1 answer:

Andrews [41]2 years ago

7 0

I think this is the graph for your answer, for the next question like this, try to use "Desmos"

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in the sixth grade 13 out of 20 students have a dog if there are 152 sixth grade students how many of them have a dog

DENIUS [597]

(13/20)=(x/152)

1976=20x

x=98.8

About 99 have dogs

4 0

3 years ago

Antonio had his hair cut at the salon. After he returned home, he realized he had accidentally tipped the stylist 114% of the co

stepan [7]

=>C

1.14*20.95=23.883

5 0

2 years ago

Solve for x, show all of your work: -2x – 4 < 10

NNADVOKAT [17]

<span>-2x – 4 < 10
add 4 to both sides
-2x - 4 + 4 < 10 + 4
simplify
-2x < 14
divide both sides by -2
-2x/-2 < 14/-2
simplify
x < -7</span>

3 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

Can somebody help me please?

madreJ [45]

Factor
(x+7)(x-2)

Proof
7-2=5
7*-2=-14

Zeros
x+7=0
x=-7

x-2=0
x=2

Final answer: C

3 0

2 years ago

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