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3 years ago

7(2x+3)=3(4x+6)+2x+3

Mathematics

1 answer:

makvit [3.9K]3 years ago

5 0

Answer:

$7(2x + 3) = 3(4x + 6) + 2x + 3 \\ 14x + 21 = 12x + 18 + 2x + 3 \\ 14x - 14x = 21 - 21 \\ 0 = 0$

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PLEASE HELP!!! 20 POINTS!!!

nikklg [1K]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

X < 36 and x > -36<br>x < 36 or x>-36<br>x< 36 or x > 0<br>x < 36 and x > 0

hjlf

Answer:

D,x < 36 and x > 0, negative number cannot be in sq.root

3 0

3 years ago

Assuming that the equation defines x and y implicitly as differentiable functions xequalsf(t), yequalsg(t), find the slope of

Doss [256]

Answer:

$\dfrac{dx}{dt} = -8,\dfrac{dy}{dt} = 1/8\\$

Hence, the slope , $\dfrac{dy}{dx} = \dfrac{-1}{64}$

Step-by-step explanation:

We need to find the slope, i.e. $\dfrac{dy}{dx}$ .

and all the functions are in terms of $t$ .

So this looks like a job for the 'chain rule', we can write:

$\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx} -Eq(A)$

Given the functions

$x = f(t)\\y = g(t)\\$

and

$x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)$

we can differentiate them both w.r.t to $t$

first we'll derivate Eq(B) to find dx/dt

$x^3 +4t^2 = 37\\3x^2\frac{dx}{dt} + 8t = 0\\\dfrac{dx}{dt} = \dfrac{-8t}{3x^2}\\$

we can also rearrange Eq(B) to find x in terms of t , $x = (37 - 4t^2)^{1/3}$ . This is done so that $\frac{dx}{dt}$ is only in terms of t.

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\\dfrac{dx}{dt} = \dfrac{-8(3)}{3(37 - 4(3)^2)^{2/3}}\\\dfrac{dx}{dt} = -8$

now let's differentiate Eq(C) to find dy/dt

$2y^3 - 2t^2 = 110\\6y^2\frac{dy}{dt} -4t = 0\\\dfrac{dy}{dt} = \dfrac{4t}{6y^2}$

rearrange Eq(C), to find y in terms of t, that is $y = \left(\dfrac{110 + 2t^2}{2}\right)^{1/3}$ . This is done so that we can replace y in $\frac{dy}{dt}$ to make only in terms of t

$\dfrac{dy}{dt} = \dfrac{4t}{6y^2}\\\dfrac{dy}{dt}=\dfrac{4t}{6\left(\dfrac{110 + 2t^2}{2}\right)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dy}{dt} = \dfrac{4(3)}{6\left(\dfrac{110 + 2(3)^2}{2}\right)^{2/3}}\\\dfrac{dy}{dt} = \dfrac{1}{8}$

Finally we can plug all of our values in Eq(A)

but remember when plugging in the values that $\frac{dy}{dt}$ is being multiplied with $\frac{dt}{dx}$ and NOT $\frac{dx}{dt}$ , so we have to use the reciprocal!

$\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx}\\\dfrac{dy}{dx} = \dfrac{1}{8}.\dfrac{1}{-8} \\\dfrac{dy}{dx} = \dfrac{-1}{64}$

our slope is equal to $\dfrac{-1}{64}$

7 0

3 years ago

4) A light is on the top of a 12 ft tall pole and a 5 ft 6 in tall person is walking away from the pole at a rate of 2 ft/sec. a

VARVARA [1.3K]

Answer:

A) 48/13 ft/sec

B) 22/13 ft/sec

Step-by-step explanation: Given that:

A light is on the top of a 12 ft tall pole and a 5ft 6in tall person is walking away from the pole at a rate of 2 ft/sec.

A) At what rate is the tip of the shadow moving away from the pole when the person is 25 ft from the pole?

Using similar triangles, we can say:

12/L = 55/(L - x)

Cross multiply to get:

12(l - x) = 5.5l

12l - 12x = 5.5l

6.5l = 12x

x = (6.5/12)l

Taking the derivative with respect to time we get:

dx/dt = (6.5/12)dl/dt or

dl/dt = (12/6.5)(dx/dt)

Since dx/dt = 2 so

dl/dt = (12/6.5)(2)

= 24/6.5

= 48/13 ft/sec

B) At what rate is the tip of the shadow moving away from the person when the person is 25 ft from the pole ?

Subtract the rate the shadow is going from the rate the man is going. Therefore

48/13 - 2 = 22/13 ft/sec.

7 0

3 years ago

Solve the expression for a=12 , b=6 and c=3

katrin2010 [14]

Answer:

Step-by-step explanation:

$a = 12, \ b = 6, \ c = 3\\\\b^2 - c^2 + (a+ b+c) = 6^2 - 3^2 + (12 +6+3) = 36-9+21 = 48$

8 0

2 years ago

Read 2 more answers

7(2x+3)=3(4x+6)+2x+3​

7(2x+3)=3(4x+6)+2x+3