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2 years ago

Ok how do you add pictures

1 answer:

5 0

Ohh well when u type in another question it should show a pic option at the bottom once your done typing whatever is u wish to ask :)

You might be interested in

Simplify the expression 7(1 +5v) =

horsena [70]

The answer is 7 + 35v

8 0

2 years ago

Read 2 more answers

If g(x)=3f(x)-8 and f(x)=10-x to the second, then what if the value of g(-2)

babymother [125]

Answer:

Step-by-step explanation:

g(x)=3(10-x)^2-8

=3(12)^2-8

=144*3-8

=432-8

=424

7 0

2 years ago

Plzz add an explanation for it too

DIA [1.3K]

Answer:

A $0.02

Step-by-step explanation:

Let's solve how much does it cost to have a bulb on for an hour. ⇒ 120 watt = 0.1 kilowatt per hour which costs 0.20 per kilowatt per ⇒ 0.1 kilowatt per hour* 0.20 = 0.02 dollars in an hour.

3 0

2 years ago

which expression is equal to 2 square root of 28 - 5 the square root of 63? a. -11 square root of 7, b. -37 the square root of 7

adell [148]

Answer:

The correct answer is A. - 11 square root of 7.

Step-by-step explanation:

We have the expression 2 √28 - 5 √ 63 and we have to find and equal expression.

Let's simplify it:

2 √28 - 5 √ 63

2 √7 * 4 - 5 √ 9 * 7 (28 is also equal to 7 * 4 and 63 is equal to 9 * 7)

2 * 2 √7 - 5 * 3 √ 7

4 √7 - 15 √7

- 11 √7

The equal expression is A. - 11 square root of 7.

6 0

2 years ago

If the price is increasing at a rate of 2 dollars per month when the price is 10 dollars, find the rate of change of the demand.

densk [106]

Answer:

The demand reduces by $7.12 per month

Step-by-step explanation:

Given

$p\to price$

$x \to demand$

$2x^2+5xp+50p^2=24800.$

$p =10; \frac{dp}{dt} = 2$

Required

Determine the rate of change of demand

We have:

$2x^2+5xp+50p^2=24800.$

Differentiate with respect to time

$4x\frac{dx}{dt} + 5x\frac{dp}{dt} + 5p\frac{dx}{dt} + 100p\frac{dp}{dt} = 0$

Collect like terms

$4x\frac{dx}{dt} + 5p\frac{dx}{dt} = -5x\frac{dp}{dt} - 100p\frac{dp}{dt}$

Factorize

$\frac{dx}{dt}(4x + 5p) = -5(x + 20p)\frac{dp}{dt}$

Solve for dx/dt

$\frac{dx}{dt} = -\frac{5(x + 20p)}{4x + 5p}\cdot \frac{dp}{dt}$

Given that: $2x^2+5xp+50p^2=24800.$ and $p = 10$

Solve for x

$2x^2 + 5x * 10 + 50 * 10^2 = 24800$

$2x^2 + 50x + 5000 = 24800$

Equate to 0

$2x^2 + 50x + 5000 - 24800 =0$

$2x^2 + 50x -19800 =0$

Using a quadratic calculator, we have:

$x \approx -113\ and\ x\approx88$

Demand must be greater than 0;

So: $x=88$

So, we have: $x=88$ ; $p =10; \frac{dp}{dt} = 2$

The rate of change of demand is:

$\frac{dx}{dt} = -\frac{5(88 + 20*10)}{4*88 + 5*10} * 2$

$\frac{dx}{dt} = -\frac{5(288)}{402} * 2$

$\frac{dx}{dt} = -\frac{2880}{402}$

$\frac{dx}{dt} \approx -7.16$

This implies that the demand reduces by $7.12 per month

8 0

2 years ago

Ok how do you add pictures​

Ok how do you add pictures