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3 years ago

What’s the domain and range for these two?

Mathematics

1 answer:

user100 [1]3 years ago

4 0

Step-by-step explanation:

1) the domain is from -3 to 5 D [-3,5]

the range is from -5 to -2. R [-5,-2]

2)the domain is from -3 to 6 but not including 6 D [-3,6)

the range is from -5 but not including -5 to 2. R (-5,2]

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Traveling 1,200 miles in 4 hours

Dafna1 [17]

Answer:

speed is 240 miles per hour

Step-by-step explanation:

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3 years ago

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1. Find each measure. a) m20 8 R Select 110" S 8 T c)

Zepler [3.9K]

Answer:

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3 years ago

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Please please please help me. I have no idea what to do

rewona [7]

We have a bunch of rectangles making up the lateral sides of this figure. They have side length 12 and height 13.

So the area of one rectangle is 12 x 13 = 156.
We have 6 of these rectangles, so our lateral area is 6(156) = 936.

The total surface area requires that we include the area of the base. To find the area of a regular polygon we take half of the product of the apothem and perimeter.

A = (1/2)*a*p

The apothem is given to be 10.39,
and the perimeter is fairly easy to calculate:
With the side length being 12, the perimeter is then 6(12) = 72.

So then the area of our base is (1/2)*10.39*72 = 374.04

So then the total surface area is the (base area) + (lateral area),
374.04 + 936 = 1310.04

3 0

3 years ago

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

3 0

2 years ago

Could someone help? :)

cluponka [151]

The value missed is 1 min / 60 seconds

<h3>What is unit conversion?</h3>

Units is the conversion between different units of measurement for the same quantity, typically through multiplicative conversion factors.

To convert 720 seconds to hour,

First convert second to minutes divide by 60 and then again convert minutes to hour again divide by 60.

So,

720 seconds * ( 1 minutes / 60 seconds) * ( 1 hour / 60 minutes)

= 12 hours.

Learn more about this concept here:

brainly.com/question/4736731

#SPJ1

8 0

2 years ago