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3 years ago

Only need the first question!!! Help me pls!! I’ll follow you and make brain list.

Mathematics

1 answer:

Alex787 [66]3 years ago

4 0

Answer:

Step-by-step explanation:

This is hard

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G(t) = -2t + 3f(t) = 2t - 5<br> Find (g f)(t)

Tanzania [10]

Answer:

( g − f ) ( a ) = g ( a ) − f ( a )

( g −f ) ( a ) = g ( a ) − f ( a )

= − 3 a − 3 − a 2 − 5

= − a 2 − 3a − 8

Step-by-step explanation:

3 0

4 years ago

Rewrite the statement in mathematical notation. (Let y be the distance from the top of the ladder to the floor, x be the distanc

In-s [12.5K]

Answer:

$\frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}$

Step-by-step explanation:

Let L be the length of the ladder,

Given,

x = the distance from the base of the ladder to the wall, and t be time.

y = distance from the base of the ladder to the wall,

So, by the Pythagoras theorem,

$L^2 = y^2 + x^2$

$\implies L = \sqrt{y^2 + x^2}$ ,

Differentiating with respect to time (t),

$\frac{dL}{dt}=\frac{d}{dt}(\sqrt{x^2 + y^2})$

$=\frac{1}{2\sqrt{x^2 + y^2}}\frac{d}{dt}(x^2 + y^2)$

$=\frac{1}{2\sqrt{x^2 + y^2}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})$

$=\frac{1}{\sqrt{x^2 +y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Here,

$\frac{dy}{dt}=-6\text{ ft per sec}$

Also, $\frac{dL}{dt} = 0$ ( Ladder length = constant ),

$\implies \frac{1}{\sqrt{x^2 +y^2}}(x(-6)+y\frac{dy}{dt})=0$

$-6x + y\frac{dy}{dt}=0$

$y\frac{dy}{dt}=6x$

$\implies \frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}$

Which is the required notation.

8 0

3 years ago

What is the least common multiple of 8, 25, and 40?

snow_lady [41]

Answer:200

Step-by-step explanation:

5 0

2 years ago

Someone help pls!!!!!!!!!!!

Gemiola [76]

The answer is no question for you lol

7 0

3 years ago

Plisss help me in thiss

ahrayia [7]

Answer:

x=80

Step-by-step explanation:

The angle of a line is 180*.

The equation would be (x+20)+x=180

Subtract 20 from both sides: 2x=160

Divide by 2: x=80

That's your answer!

I hope this helps!

3 0

3 years ago

Read 2 more answers