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2 years ago

8

If A is directly proportional to C and B is directly proportional to C, prove that each of the following is directly proportiona

l to C.
(a) A + B

(b) A-B

(c) √AB

1 answer:

Yuliya22 [10]2 years ago

7 0

Answer:

A+B is correct answer I hope it helps you

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subset to real numbers, and as you can see are not at all connected to the rest of numbers (and integers).

Natali5045456 [20]

Irrational numbers are the subset of real numbers that are not at all connected to the rest of numbers.

The subsets of real numbers are natural numbers, whole numbers, integers, rational numbers and irrational numbers.

Natural numbers are subset of whole numbers, which are subset of integers, which are subset of rational numbers. Hence, all of them are interconnected. The set of irrational numbers is the only subset of real numbers which is not associated with the rest.

For example:-

1 is a natural number, whole number, integer, rational number but not irrational number. On the other hand, $\sqrt{2}$ is an irrational number but none of the rest.

To learn more about real numbers, here:-

brainly.com/question/551408

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1 year ago

if you deposit $860 in a new account that earns 6.5% simple interest. After 6 months, how much interest will you have learned?

pishuonlain [190]

860*0.0325 = 27 dollars and 95 cents

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3 years ago

If each pound of beef costs $3.69, how much does 2 pounds cost? Round to the nearest cent.

Kryger [21]

Answer: Two pounds of beef costs $7.38.

Step-by-step explanation:

Multiply the cost for one pound by the amount of pounds you need.

$3.69×2=$7.38

I hope this helps!!

6 0

1 year ago

Read 2 more answers

Answer the following questions about 2 numbers chosen from the numbers 1 through 12 (including 1 and 12): Suppose the numbers ar

Nana76 [90]

Forts then you take the 1-6 to get the money back and get your credit

3 0

2 years ago

Explain how to find the vertex of 2x^2+12x+16

Elina [12.6K]

The vertex form:

$y=a(x-h)^2+k\\\\(h,\ k)-vertex$

Use $(a+b)^2=a^2+2ab+b^2\qquad(*)$

$2x^2+12x+16=2(x^2+6x)+16=2(x^2+2(x)(3))+16\\\\=2(\underbrace{x^2+2(x)(3)+3^2}_{(*)}-3^2)+16=2((x+3)^2-9)+16\\\\=2(x+3)^2+(2)(-9)+16=2(x+3)^2-18+16\\\\=\boxed{2(x-(-3))^2-2}\to h=-3,\ k=-2$

<h3>Answer: (-3, -2).</h3>

Other method:

$f(x)=ax^2+bx+c\\\\vertex:(h,\ k)\\\\h=\dfrac{-b}{2a},\ k=f(h)$

We have

$f(x)=2x^2+12x+16\\\\a=2,\ b=12,\ c=16$

Substitute:

$h=\dfrac{-12}{(2)(2)}=\dfrac{-12}{4}=-3\\\\k=f(-3)=2(-3)^2+12(-3)+16=2(9)-36+16=18-36+16=-2$

<h3>Answer: (-3, -2).</h3>

5 0

3 years ago