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3 years ago

Which conjunction about whole numbers is true?

Mathematics

2 answers:

alexandr402 [8]3 years ago

6 0

Answer:

The sum of two even numbers is even and greater than the addends.

Step-by-step explanation:

Let 2n and 2m be two even numbers.

The sum of these numbers is

$2n+2m$

Use distributive property:

$2(n+m)$

This number is even because 2 is its factor.

Note that

$2n+2m>2n\\ \\2n+2m>2m$

Thus, correct option is:

The sum of two even numbers is even and greater than the addends.

GaryK [48]3 years ago

4 0

Answer:

Step-by-step explanation:

Just took the test

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Read 2 more answers

You have$200in a savings account and $100 in a checking account. Each week you add $50 to your saving account and $30 to your ch

enyata [817]

Answer:

The expression is (80w+300)

Step-by-step explanation:

Let

y ----> the total amount of money in both accounts

s ----> the amount of money in saving account

c ----> the amount of money in checking account

w ----> the number of weeks

we know that

saving account

s=50w+200 -----> equation A

checking account

c=30w+100 ----> equation B

To find out the total amount of money in both accounts, sum equation A and equation B

y=s+c

substitute

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therefore

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5 0

3 years ago

A Norman window is a window with a semi-circle on top of regular rectangular window. (See the picture.) What should be the dimen

Vikki [24]

Answer:

bottom side (a) = 3.36 ft

lateral side (b) = 4.68 ft

Step-by-step explanation:

We have to maximize the area of the window, subject to a constraint in the perimeter of the window.

If we defined a as the bottom side, and b as the lateral side, we have the area defined as:

$A=A_r+A_c/2=a\cdot b+\dfrac{\pi r^2}{2}=ab+\dfrac{\pi}{2}\left (\dfrac{a}{2}\right)^2=ab+\dfrac{\pi a^2}{8}$

The restriction is that the perimeter have to be 12 ft at most:

$P=(a+2b)+\dfrac{\pi a}{2}=2b+a+(\dfrac{\pi}{2}) a=2b+(1+\dfrac{\pi}{2})a=12$

We can express b in function of a as:

$2b+(1+\dfrac{\pi}{2})a=12\\\\\\2b=12-(1+\dfrac{\pi}{2})a\\\\\\b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a$

Then, the area become:

$A=ab+\dfrac{\pi a^2}{8}=a(6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a)+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a^2+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)a^2\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right)a^2$