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AlekseyPX
3 years ago
7

Write a decimal that is 1/10 of 30

Mathematics
1 answer:
OLEGan [10]3 years ago
3 0

\dfrac{1}{10}\ of\ 30\to\dfrac{1}{10}\cdot30=0.1\cdot30=3.0

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What is true about the completely simplified sum of the polynomials 3x squared y squared minus 2 x y to the power of 5 and negat
12345 [234]

Answer:

The sum is a binomial with a degree of 6.

Step-by-step explanation:

To find : What is true about the completely simplified sum of the polynomials 3x^2y^2-2xy^5 and -3x^2y^2+3x^4y ?

Solution :

The polynomials  3x^2y^2-2xy^5 and -3x^2y^2+3x^4y ,

The sum of polynomial is given by,

S=3x^2y^2-2xy^5+(-3x^2y^2+3x^4y)

On simplifying , we get,

S=3x^2y^2-2xy^5-3x^2y^2+3x^4y

S=-2xy^5+3x^4y

Therefore, the result obtained is a binomial with a degree of 6.

4 0
3 years ago
A bond quoted at 93 (1/8) is equal to ___ per 1,000.00 of the face amount.
Ksju [112]
The number a bond is the percentage of the value of the bond that the bond is worth. Because the bond is quoted at 93, the bond is worth $930 per $1,000.
5 0
3 years ago
Read 2 more answers
Enrico deposited $2000, in a savings account. Each month he will deposit an additional $25. Which kind of function best models t
AfilCa [17]

Answer:

<em> Linear model </em>

The function f(x) which represents the total amount in the savings account in x months is given by:           f(x)=25x+2000

Step-by-step explanation:

Given:

Enrico deposited $2000 in a savings account.

Each month he will deposit additional $25.

This shows that the rate at which the amount is increasing each month is constant.

Therefore, the model will be  linear with a slope 25.

So , if x represents the number of months.

and f(x) represents the corresponding amount in the account.

Then the function f(x) is given by: f(x)=25x+2000

3 0
3 years ago
Forty children from an after-school club went to the matinee.This is 25% of the children in the club.How many children are in th
Korvikt [17]
25% = 0.25

40 / 0.25 = 160

40 is 25% of 160.

There are 160 children in the club.
7 0
3 years ago
Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
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