All categories

3 years ago

NEED ANSWERS NOW

Mathematics

1 answer:

IrinaK [193]3 years ago

3 0

The answer is choice C

On the left side we have something in the form a*b+a*c+a*d

On the right side we have something in the form a*(b+c+d)

All together, we have a*b+a*c+a*d = a*(b+c+d), where the letters a,b,c,d are to be replaced with the proper values (0.5, 0.5, 0.3, and 0.2 in that exact order).

The idea of distribution is to multiply the outer value 0.5 by each of the values inside the parenthesis (0.5, 0.3, and 0.2) and add up the products.

You might be interested in

Use an Addition or Subtraction Formula to find the exact value of the expression, as demonstrated in Example 1. cos 41π 12.

PtichkaEL [24]

Answer: $\frac{\sqrt{6}-\sqrt{2}}{2}$

Step-by-step explanation:

We apply the formula $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$ .

Note that $\cos(\frac{41}{12}\pi)=\cos((\frac{36}{12}+\frac{7}{12})\pi)=\cos(3\pi + \frac{7}{12})\pi)$ . Take $x=3\pi$ and $y=\frac{7}{12}\pi$ in the formula above to get

$\cos(\frac{41}{12}\pi)=\cos(3\pi)\cos(\frac{7}{12}\pi)-\sin(3\pi)\sin(\frac{7}{12}\pi)=(-1)\cdot \cos(\frac{7}{12}\pi)-0\cdot\sin(\frac{7}{12}\pi)=-\cos(\frac{7}{12}\pi)$

Then the value of this expression is $-\cos(\frac{7}{12}\pi)$

We can use the cosine addition formula again to simplify further. Decompose the fraction in the argument as:

$\cos(\frac{7}{12}\pi)=\cos((\frac{3}{12}+\frac{4}{12})\pi)=\cos((\frac{1}{4}\pi + \frac{1}{3})\pi)$

Applying the formula with $x=\frac{1}{4}\pi$ and $y=\frac{1}{3}\pi$ we obtain

$\cos(\frac{7}{12}\pi)=\cos(\frac{1}{4}\pi)\cos(\frac{1}{3}\pi)-\sin(\frac{1}{4}\pi)\sin(\frac{1}{3}\pi)=\frac{\sqrt{2}}{2}\cdot\frac{1}{2} -\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{2}-\sqrt{6}}{2}$

We conclude that this expression has the value $-\frac{\sqrt{2}-\sqrt{6}}{2}=\frac{\sqrt{6}-\sqrt{2}}{2}$

8 0

3 years ago

I REALLY NEED HELP ON THIS PLS HELP

Eva8 [605]

Part (a)

Answer: See the attached image below to see the filled out chart.

Note how rational and irrational numbers have nothing in common. This means there is no overlap. So they go in the rectangles. The two sets of numbers join up to form the entire set of real numbers.

Integers are in the set of rational numbers. This is because something like 7 is also 7/1; however 1/7 is not an integer. So not all rational numbers are integers. The larger purple circle is the set of integers.

The smaller blue circle is the set of whole numbers. The set of whole numbers is a subset of integers. Recall the set of whole numbers is {0,1,2,3,...} so we ignore the negative values only focusing on 0 and positive numbers that don't have any fractional values. In contrast, the set of integers is {..., -3, -2, -1, 0, 1, 2, 3, ...} here we do include the negatives.

============================================================

Part (b)

1) Some irrational numbers are integers.

This is false. An irrational number is not rational. The set of integers is contained entirely in the set of rational numbers.

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

2) Some whole numbers are not irrational numbers.

This is false. The statement implies that some whole numbers are irrational, but the set of whole numbers is inside the set of rational numbers, which has no overlap with the irrationals. The statement should be "All whole numbers are not irrational numbers".

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

3) All rational numbers are whole numbers.

This is false. A rational number like 1/3 is not a whole number.

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

4) All integers are whole numbers

This is false. An integer like -44 is not a whole number because the set of whole numbers is {0,1,2,3,...} and we're not including negative values.

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

In summary, all four statements for part (b) are false.