All categories

3 years ago

What are the equations that have the same solutions as 3x-12=24 between

Mathematics

2 answers:

gayaneshka [121]3 years ago

6 0

C because
3x-12=24
+12+12
_________
3x=36

Effectus [21]3 years ago

5 0

Answer:

ddqsc

Step-by-step explanation:

rdfgrrffferrrrrddr

You might be interested in

If you have two pennies, the probability of both coming up heads when flipped is _____.

aleksandr82 [10.1K]

one penny has a 50% chance for heads or tails so if you add one more penny it drops the chances for two heads by 50% again so in the ende you only have 1/4 the chancre of two heads up. hope it helps! :)

4 0

3 years ago

Read 2 more answers

What is the slope of the line that passes through the points (-4, 2) & (1,5)

vladimir2022 [97]

Answer:

Step-by-step explanation:

Slope= y2-y1/x2-x1

5-2/1-(-4) = 3/5

Slope: 0.6

4 0

3 years ago

I-6I = <br><br> (whats the answer and show your work)

Tomtit [17]

Answer:

um i think it's 5l

cause l=1

4 0

3 years ago

Read 2 more answers

How to solve part ii and iii

iragen [17]

(i) Given that

$\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}$

when $x=1$ this reduces to

$\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}$

$\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}$

$2 \tan^{-1}(y) = \dfrac\pi3$

$\tan^{-1}(y) = \dfrac\pi6$

$\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)$

$\implies \boxed{y = \dfrac1{\sqrt3}}$

(ii) Differentiate $\tan^{-1}(xy)$ implicitly with respect to $x$ . By the chain and product rules,

$\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}$

(iii) Differentiating both sides of the given equation leads to

$\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0$

where we use the result from (ii) for the derivative of $\tan^{-1}(xy)$ .

Solve for $\frac{dy}{dx}$ :

$\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0$

$\left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} = -\left(\dfrac1{1+x^2} + \dfrac y{1+x^2y^2}\right)$

$\dfrac{1+x^2y^2 + x(1+y^2)}{(1+y^2)(1+x^2y^2)} \dfrac{dy}{dx} = - \dfrac{1+x^2y^2 + y(1+x^2)}{(1+x^2)(1+x^2y^2)}$

$\implies \dfrac{dy}{dx} = - \dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2) (1 + x^2y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2) (1+x^2y^2)}$

$\implies \dfrac{dy}{dx} = -\dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2)}$

From part (i), we have $x=1$ and $y=\frac1{\sqrt3}$ , and substituting these leads to

$\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}$

$\dfrac{dy}{dx} = -\dfrac{\left(\frac43 + \frac2{\sqrt3}\right) \times \frac43}{\frac83 \times 2}$

$\dfrac{dy}{dx} = -\dfrac13 - \dfrac1{2\sqrt3}$

as required.

3 0

2 years ago

Which of the following is the best example of something considered multimedia.

Natalka [10]

Answer: a collage using paper, wood, and paint

Step-by-step explanation:

3 0

3 years ago