Ask question

Ask question

All categories

3 years ago

11

0.92-0.51 plz help i dnt undersatnd

2 answers:

Mashutka [201]3 years ago

4 0

The answer is 0.41 but what do you not understand? Hope I helped!

AveGali [126]3 years ago

3 0

0.92 - 0.51 = 0.41.

0.92
0.51 -
--------
0.41

You might be interested in

What is the truth value for the following conditional statement?

Studentka2010 [4]

Answer:

TT→T

Step-by-step explanation:

If p is false, then ~p is true.

If q is false, then ~q is true.

Now note that

If a and b are both true, then a→b is true.
If a is true, b is false, then a→b is false.
If a is false, b is true, then a→b is true.
If a and b are both false, then a→b is true.

In your case, both~p and ~q are true, then ~p→~q is true too (or TT→T)

3 0

3 years ago

What refers to the quantity of goods and services that consumers are willing to buy at a given price?

crimeas [40]

Answer:

"demand"

Step-by-step explanation:

Vocabulary question.

"Demand" refers to the quantity of goods and services that consumers are willing to buy at a given price.

4 0

2 years ago

8. In a circle of any size, what ratio does

ahrayia [7]

The answer should be C

3 0

2 years ago

Describe how you would determine the x-intercepts from the factored form of a quadratic equation

Aloiza [94]

Answer:

Please check explanations

Step-by-step explanation:

Given the factored form, all we have to do is to equate each of the factor to zero, then solve for x

The particular value of x represents the solution that shows the value of the x-intercept

for example

In a case of a quadratic equation where we have

(x-a)(x-b) = 0

Then we can have

(x-a) = 0 or ((x-b) = 0

Then we have x = a or x = b

a and b are the x intercepts in this case

4 0

2 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

1 year ago