All categories

2 years ago

What is the value of the number in the thousandths place 7.819

Mathematics

2 answers:

MrMuchimi2 years ago

8 0

Answer:

the value of the number in the thousandths place is 9

Step-by-step explanation:

No calculator!

kvasek [131]2 years ago

5 0

9 is in the thousandths place

You might be interested in

(32 +1) + (5 + 7z) = Help please

levacccp [35]

Answer: -5.428

Step-by-step explanation:

33 + 5 + 7z = 0

38 + 7z = 0

7z = -38

z = -38/7

5 0

2 years ago

What is the sum of the measures of the exterior angles of this triangle?

Serjik [45]

the sum of the measures of the exterior angles of this triangle is 360.
(A=61, B=119, C=68 in case you need to know)

8 0

2 years ago

Read 2 more answers

Is the square root of 3,600 rational or irrational? Explain.

Daniel [21]

Answer:

Rational. the square root of 3600 is a real, rational number, 60.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Mathmatical proofs. please help

belka [17]

Answer looking for the part of the proof that is wrong
They switched the right angles with their opposite
See picture, that’s the best way I could show you without typing the whole thing over
Switch the two Right angles in the sentence

6 0

1 year 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

3 years ago