All categories

3 years ago

How do I add: 1/3+1/4?

Mathematics

2 answers:

vlada-n [284]3 years ago

6 0

You convert the denominator into a common number. That is 12. You then add 4/12 + 3/12 = 7/12

inn [45]3 years ago

3 0

Get a common factor which would be 12.... the equation is now 4/12 + 3/12 which is 7/12

You might be interested in

Not rlly a paper I gotta do buuuut, say I have seven one hundreds, one ninety nine, and two zeros... what would my grade be in t

vlabodo [156]

If they’re all weighted equally, then your average grade will be 79.9%

8 0

3 years ago

Which of the following have the property that a(x)=a−1(x)? I. y=x II. y=1/x III.y=x^2 IV. y=x^3 A. I and II, only B. IV, only C.

valentina_108 [34]

Answer:

Correct answer:

A. I and II

Step-by-step explanation:

First of all, let us have a look at the steps of finding inverse of a function.

1. Replace y with x and x with y.

2. Solve for y.

3. Replace y with $f^{-1}(x)$

Given that:

$I.\ y=x \\II.\ y=\dfrac{1}x \\III.\ y=x^2 \\IV.\ y=x^3$

Now, let us find inverse of each option one by one.

I. y = x, a(x) = x

Replacing y with and x with y:

x = y

x = $a^{-1}(x)$ = $a(x)$ Hence, I is true.

II. $y =\dfrac{1}{x}$

Replacing y with and x with y:

$x =\dfrac{1}{y}$

$x=\dfrac{1}{a^{-1}(x)}$

$\Rightarrow a^{-1}(x) = \dfrac{1}{x}$

$a^{-1}(x)$ = $a(x)$ Hence, II is true.

III. $y =x^{2}$

Replacing y with and x with y:

$x =y^{2}\\\Rightarrow y = \sqrt x\\\Rightarrow a^{-1}(x) = \sqrt{x} \ne a(x)$

Hence, III is not true.

IV. $y =x^{3}$

Replacing y with and x with y:

$x =y^{3}\\\Rightarrow y = \sqrt[3] x\\\Rightarrow a^{-1}(x) = \sqrt[3]{x} \ne a(x)$

Hence, IV is not true.

Correct answer:

A. I and II

4 0

3 years ago

Help me pls with my mathh

rjkz [21]

Sure let’s see.... I believe it is D, but I’m not 100%. Don’t give low ratings. I tried.

7 0

3 years ago

IS 10×1000 a multiplication fact?

Tanya [424]

Yes if u can multiply 10 by 1000 then it is a multiplication fact

4 0

3 years ago

Read 2 more answers

Find a function ω(x) such that the function f(x) = sin(ω(x)) has the roots at π, π2

atroni [7]

So, I came up with something like this. I didn't find the final equation algebraically, but simply "figured it out". And I'm not sure how much "correct" this solution is, but it seems to work.

$f(x)=\sin(\omega(x))\\\\f(\pi^n)=\sin(\omega(\pi^n))=0, n\in\mathbb{N}\\\\\\\sin x=0 \implies x=k\pi,k\in\mathbb{Z}\\\Downarrow\\\omega(\pi^n)=k\pi\\\\\boxed{\omega(x)=k\sqrt[\log_{\pi} x]{x},k\in\mathbb{Z}}$

8 0

3 years ago