All categories

3 years ago

How multiplied whole number with fraction

2 answers:

Volgvan3 years ago

7 0

I think it helps to turn the whole number into a fraction, just by adding a denominator of 1. Then, you can easily multiply across with both fractions, and simplify the answer once done. Hope this helps!

Fofino [41]3 years ago

3 0

Search it up and you shall find an answer believe me it is there

You might be interested in

A home improvement store buys snow shovels from a supplier for $8.50. The day before a snowstorm the store manager marks up the

balandron [24]

Answer:

The retail price is $15.30

Step-by-step explanation:

To solve this, turn the percentage into a decimal. Do this by dividing it by 100.

180 ÷ 100 = 1.8

Then, multiply the supplier price by the decimal

8.5 × 1.8 = $15.30

7 0

2 years ago

Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m < 20

Vsevolod [243]

Answer:

18 values for n are possible.

Step-by-step explanation:

Given the quadratic polynomial:

$x^2 - mx + n$

such that:

Roots are positive prime integers and

$m < 20$

To find:

How many possible values of $n$ are there ?

Solution:

First of all, let us have a look at the sum and product of a quadratic equation.

If the quadratic equation is:

$Ax^{2} +Bx+C$

and the roots are: $\alpha$ and $\beta$

Then sum of roots, $\alpha+\beta = -\frac{B}{A}$

Product of roots, $\alpha \beta = \frac{C}{A}$

Comparing the given equation with standard equation, we get:

A = 1, B = -m and C = n

Sum of roots, $\alpha+\beta = -\frac{-m}{1} = m$

Product of roots, $\alpha \beta = \frac{n}{1} = n$

We are given that $m$

$\alpha$ and $\beta$ are positive prime integers such that their sum is less than 20.

Let us have a look at some of the positive prime integers:

2, 3, 5, 7, 11, 13, 17, 23, 29, .....

Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.

So, possible combinations and possible value of $n (= \alpha \times \beta)$ are:

$1.\ 2, 2\Rightarrow n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow n = 6\\3.\ 2, 5 \Rightarrow n = 10\\4.\ 2, 7\Rightarrow n = 14\\5.\ 2, 11 \Rightarrow n = 22\\6.\ 2, 13 \Rightarrow n = 26\\7.\ 2, 17 \Rightarrow n = 34\\8.\ 3, 3\Rightarrow n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow n = 15\\10.\ 3, 7 \Rightarrow n = 21\\$

$11.\ 3, 11\Rightarrow n = 33\\12.\ 3, 13 \Rightarrow n = 39\\13.\ 5, 5 \Rightarrow n = 25\\14.\ 5, 7 \Rightarrow n = 35\\15.\ 5, 11 \Rightarrow n = 55\\16.\ 5, 13 \Rightarrow n = 65\\17.\ 7, 7 \Rightarrow n = 49\\18.\ 7, 11 \Rightarrow n = 77$

So,as shown above 18 values for n are possible.

3 0

3 years ago

In a geometric sequence, a2=2, a3=16, and a4=128.

Oduvanchick [21]

Answer:

an = 1/4(8)^n-1

Step-by-step explanation:

Given the following in a geometric sequence, a2=2, a3=16, and a4=128.

nth term of a sequence = ar^n-1

a is the first term

r is the common ratio

r = a3/a2 = a4/a3

r = 16/2 = 128/16

r = 8

a2 = ar

2 = 8a

a = 2/8

a = 1/4

The nth term by substituting the parameters will be;

an = 1/4(8)^n-1

4 0

3 years ago

How many milligrams of medication are in 75 mL of a 2% solution? Help

deff fn [24]

The answer is 23 I took the test

4 0

3 years ago

Which of the following BEST summarizes how to use the multiplication property of equality to isolate the

borishaifa [10]

Answer:

A. divide only on the right side of the equation by 2.

Step-by-step explanation:

I took the test yesterday and I got it hope this helps.

7 0

2 years ago