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2 years ago

11

Part a: what is the initial value of the function? what does it mean in the context of the situation?

1 answer:

ser-zykov [4K]2 years ago

4 0

Answer:

Initial value = 10, slope = 4

Step-by-step explanation:

Initial value is the point where x takes its lowest value. In this case it is x = 0 and y = 10. Rate of change is the slope at a given point. Because this is a first order equation graph the slope is always the same. And in this case it is 4 as for every 1 unit of x the y goes 4.

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What is 144 divided by 3 and does it have a remainder

Karolina [17]

The answer is 48 and no.

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3 years ago

Read 2 more answers

The scale of a map is 1 inch = 60 miles. Find the actual distance if the distance on the distance on the map is 3 1/2 inches.

laila [671]

Answer:

210 miles

Step-by-step explanation:

The scale is

1 in = 60 miles

Find the distance of equivalent to 3 1/2 in

3 1/2 * 60 miles =
3*60 + 1/2*60 miles =
180 + 30 miles =
210 miles

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1 year ago

Find m<1.<br> A.30<br> B.35<br> C.40<br> D.55

mihalych1998 [28]

Answer:

m <1 = 30

Step-by-step explanation:

The sum of the angles of a triangle add to 180

Let x be the unknown angle

75+75+x = 180

Combine like terms

150+x =180

Subtract 150 from each side

150-150+x = 180-150

x = 30

5 0

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.

6 0

2 years ago

What is the volume of the following rectangular prism?<br> 21/2 units <br> 8/3 units

xeze [42]

Answer:

28 units³

Step-by-step explanation:

The volume (V) of the prism is calculated as

V = Al ( A is the area of end face and l is the length )

Here A = $\frac{21}{2}$ and l = $\frac{8}{3}$ , thus

V = $\frac{21}{2}$ × $\frac{8}{3}$ ( cancel 21 and 3 by 3 and 2 and 8 by 2 )

= $\frac{7}{1}$ × $\frac{4}{1}$ = 7 × 4 = 28 units³

5 0

3 years ago