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3 years ago

5

The perimeter of a right triangle is 24 in. find the dimensions of the triangle if the aide are all whole-number lengths.

1 answer:

Tems11 [23]3 years ago

7 0

it can be 8 on each side or 7 9 and 8 or any 3 numbers you can add together to get 24.

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Please just answer don’t ask

frozen [14]

Answer:

20

The strawberry is 3, The watermelon is 3, The peach is 9, and the other one is 5

3+3+9+5=20

5 0

3 years ago

Read 2 more answers

Simplify the expression i⁸²

Elza [17]

Answer:

We can use a trick here. Let's look at the first few exponents of i to realize this:

i^0 = 1

i^1 = i

i^2= -1

i^3 = -i

i^4 = 1

i^5 = i

i^6 = -1

i^7 = -i

we can see that the pattern (1, i, -1, -i) repeats. Since 82/2 = 41, and 41 is only divisible by 1, i^41 = i, and i^2 = -1. -1*i = -i, so i^82 = -i.

Step-by-step explanation:

6 0

3 years ago

A car travels at a constant speed.

ad-work [718]

Answer:

see below

Step-by-step explanation:

<h3>Given</h3>

Distance is 142.2 m, correct to 1 decimal place
Time is 7 seconds, correct to nearest second

<h3>To find:</h3>

Upper bound for the speed

<h3>Solution </h3>

<em>Upper bound for the speed = upper bound for distance/lower bound for time</em>

Upper bound for distance = 142.25 m (added 0.1/5 = 0.05)
Lower bound for time = 6.5 seconds (subtracted 1/2 = 0.5)

<u>Then, the speed is:</u>

142.25/6.5 = 21.88 m/s
21.88 = 21.9 m/s correct to 1 decimal place
21.88 = 22 m/s correct to nearest m/s

4 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

75% of what number is 45? <br><br> A:15<br> B:30<br> C:45<br> D:60

jarptica [38.1K]

Answer:

D. 60= 100 % therefore, 45 would equal 75 %.The answers D. Hope this helps!

Step-by-step explanation:

5 0

3 years ago