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3 years ago

How do I find c and e?

Mathematics

1 answer:

sdas [7]3 years ago

4 0

I dont know with the question your asking but I would maybe compare it with the 8 and 6 sorry I couldn't help much!

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If y+(x^2+x+1)(x^3-x-1), then y’(2)=

umka2103 [35]

Easy it’s y=learn it your self and pay attention you bum

3 0

3 years ago

A family has two cars. The first car has a fuel efficiency of 25 miles per gallon of gas and the second has a fuel efficiency of

Oxana [17]

first car is x

second car is y

25x+30y=1550

x+y=55

Multiply the second equation by -25 for elimination method to cancel out x

-25x-25y=-1375

Add the two equations

5y=175

y=35

Plug this value into the second equation

x+35=55

x=20

Final answers:

First car=20 gallons

Second car=35 gallons

8 0

3 years ago

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

Rahul has 5 BLUE BLOCKS, 3 PINK BLOCKS AND 4 GREEN BLOCKS. He wants to count the total number of blocks he is having. In how man

arsen [322]

Answer:

a) He can find the answer in one way

b) We make use of Addition property of mathematics

c) Total number of blocks = 12 blocks

Step-by-step explanation:

a) He can find the answer in one way

b) We make used of addition property of mathematics

c) The total number of blocks he is having is calculated as:

Blue blocks + Pink blocks + Green blocks

Where

Blue blocks = 5

Pink blocks = 3

Green blocks = 4

Hence:

Total number of blocks = 5 + 3 + 4 = 12 block

6 0

3 years ago

Help please.. its quiet easy anyway xd

padilas [110]

Answer:

An equilateral triangle is a triangle in which all three sides are equal. An equilateral triangle is also equiangular, in which all three internal angles are also congruent to each other and are each 60°.

Parallelogram and rhombus possibly.

Trapezoid.

8 0

4 years ago

Read 2 more answers