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3 years ago

What is the solution to the inequality? what is the solution to be mainly -90 – 15 <3

Mathematics

1 answer:

alekssr [168]3 years ago

5 0

Answer: x>−2

Step-by-step explanation:

−9x−15<3

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M+5n=p for m <br> Solving for a variable

Verdich [7]

Answer:

M=p-5n

Step-by-step explanation:

Here we are given that M+5n=p

We are asked to solve this equation for M. In order to do that we will follow these steps.

Subtracting 5n from both sides we get

M+5n-5n=p-5n

M=p-5n

Hence , now we have our M as dependent variable, whose value depends on the values of p and n. One or both of them can be independent variable/.

5 0

4 years ago

Isaac take 5 minutes to ride his bike to down the hill to school and 10 minutes to ride up the hill from school. He attends scho

Delvig [45]

In the given question, it is given that

Isaac take 5 minutes to ride his bike to down the hill to school and 10 minutes to ride up the hill from school. He attends school Monday through Friday.

So per day, from Monday to Friday, Issac rides 15 minutes.

And in 2 weeks, Issac have to ride 10 days.

So in 10 days, Issac rides,

$10*15 = 150 minutes$

7 0

3 years ago

Please help me on this math problem

Fed [463]

We know the first term is 16, because f(1) = 16. Next, we can use the recursive rule provided. f(3)=1/2*f(3-1). This gives us f(3)=1/2*f(2). We know f(2) is 8. We know have f(3)=1/2*8. This is 4.

The answer is 16, 4.

7 0

3 years ago

Find the differential of each function. (a) y = tan( 5t ) dy = Correct: Your answer is correct. (b) y = 5 − v2 5 + v2

Trava [24]

Answer:

$(a)$ $dy = 5sec^2(5t) \ dt$

$(a) \ dy = \frac{-20v}{(5+v^2)^2} \ dt$

Step-by-step explanation:

Given;

(a) y = tan(5t)

$(b) \ y = \frac{5-v^2}{5+v^2}$

Solving for (a)

y = tan(5t)

let u = 5t

⇒y = tan(u)

du/dt = 5

dy/du = sec²u

$\frac{dy}{dt} =\frac{dy}{du} *\frac{du}{dt} \\\\\frac{dy}{dt} = sec^2(u)*5\\\\\frac{dy}{dt} = 5sec^2(u)\\\\\frac{dy}{dt} = 5sec^2(5t)$

$dy = 5sec^2(5t) \ dt$

Solving for b;

let u = 5 - v²

du/dv = -2v

let v = 5+ v²

dv/du = 2v

$\frac{dy}{dv} = \frac{vdu - udv}{v^2} \\\\\frac{dy}{dv} = \frac{-2v(5+v^2) - 2v(5-v^2)}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-10v-2v^3-10v+2v^3}{(5+v^2)^2}\\\\$

$\frac{dy}{dv} = \frac{-10v-10v}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-20v}{(5+v^2)^2}\\\\dy = \frac{-20v}{(5+v^2)^2} dt$

4 0

3 years ago

F(x)=x^2. What is g(x)?

mezya [45]

Answer:

Step-by-step explanation:

TRUST.

if im wrong im really sorry

i would have failed you

i would feel guilty

and I would forget about it in like 5 minutes

6 0

3 years ago