How do you multiply a decimal by a fraction

Answers

Answer 1
Answer: In order to multiply a fraction and a decimal, you have to either change the fraction to a decimal or change the decimal to a fraction then multiply ... for example

1/2 × .8 =

we can change 1/2 to a decimal by doing the division that is described by the fraction... the fraction bar literally means "divided by", so 1/2 means 1 ÷ 2 = .5

Now you can multiply
.8 × .5 = .40 or .4

The other choice would be to convert .8 into a fraction. Look at the place value... the 8 is in the 10ths place so the value is 8 tenths or 8/10. Now multiply
8/10 × 1/2 = 8/20 = 4/5

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What is 35 5/7 multiplied by 3.14 or multiplied by 22/7

Answers

The answer for 35 5/7 x 3.14 is 112.14
The answer for 35 5/7 x 22/7 is 112.24

Answer:

112 and 12/49

Step-by-step explanation:

I'm not really sure if this answer is accurate but hope it helps somewhat. This is the answer for 35 5/7 x 22/7

PLEASE HELP QUICKLY, WILL MARK BRAINLIEST, 50 PTSAnother way to present the Fundamental Theorem of Algebra is to state that, "For any polynomial equation of degree n with real coefficients, there exist n complex solutions." Why is it important to learn about complex solutions? Why is it remarkable that with real coefficients, there must be complex solutions?

Answers

Answer:

it is important to learn about complex solution because polynomial equation formed over a complex number can only be solved by a complex number.

this is so because, the fundamental theorem of algebra state that every polynomial equation in one variable with complex coefficient has at least one complex solution.

Step-by-step explanation:

it is important to learn about complex solution because polynomial equation formed over a complex number can only be solved by a complex number.

this is so because, the fundamental theorem of algebra state that every polynomial equation in one variable with complex coefficient has at least one complex solution.

for example:

Given any positive integer n ≥ 1 and any choice of complex numbers a0,a1,...,an, such that an 6= 0,

the polynomial equation

anzn +···+ a1z + a0 = 0 (1) has at least one solution z ∈C.

No analogous result holds for guaranteeing that a real solution exists to Equation (1) if we restrict the coefficients a0,a1,...,an to be real numbers.

E.g., there does not exist a real number x satisfying an equation as simple as x2 + 1 = 0. Similarly, the consideration of polynomial equations having integer (resp. rational) coefficients quickly forces us to consider solutions that cannot possibly be integers (resp. rational numbers).

NEED AN ANSWER ASAP!!! Will Mark Brainliest!!!
Select the graph of the solution. Click until the correct graph appears. |x| = 1

Answers

A little confused by your question, but it might go something like this?

1 < x <  - 1

Answer:

the one with no arrows and only two dots


Step-by-step explanation:


Michelle went on a hike she started at 6:45am and returned at 3:28pm. How long did she hike

Answers

9 hours and 15 minutes
9 hours and 15 minutes

This scatter plot for a fund-raiser shows the time spent raising money for 9 student events. Noting the trend in the scatter plot, which is the most reasonable approximation of the money raised in the 11-hour fund-raiser?

Answers

Answer:

About $90, because the trend line shows that value for $11.

Step-by-step explanation:

It is the only one that shows a constant rate of change.

Answer:

the first one ( the line closest to 90)

Step-by-step explanation:

Round 43,000 to the nearest thousand

Answers

43,000 is the answer, you don't have to round.