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3 years ago

I am a number between 60 and 100 my ones digit is two less than the digit in my ten. I'm a prime number what is my number?

Mathematics

1 answer:

Andru [333]3 years ago

6 0

The only possibility is 79 hope it helped:D

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Which equation represents the line that has a slope of 4 and passes through the point (2, 7)?

Wittaler [7]

Answer:

The answer is y = 4x - 1.

Step-by-step explanation:

☆The equation we'll be using: y = mx + b

☆Since we already have the slope. we have to find the y-intercept.

☆ $y = mx + b \\ \\ 7 = 4(2) + b \\ 7 = 8 + b \\ \frac{ - 8 = - 8 \: \: \: \: \: \: \: \: \: \: \: }{ - 1 = b \: \: \: \: \: \: \: \: \: \: }$

☆Using the information we gathered and was given, we just to put them together.

☆ $y = 4x - 1$

3 0

3 years ago

Amelia rented a dvd and it was due to be returned on 26 november

SashulF [63]

Answer:

Step-by-step explanation:

I'm not familiar with the conversions from American dollars. I can help you solve it though.

Count how many days from Nov 26 to Dec 26 and multiply that by 9p.

It's 30 days by 9p. Take that total and convert it.

Hopefully this helps you.

8 0

3 years ago

For questions 1 – 6, find the area of the circle to the nearest hundredth.<br> Any help appriciated!

Ganezh [65]

Step-by-step explanation:

The area of a circle is given by :

A = πr²

r is the radius of the circle

(4) Diameter = 22 cm

Since, radius = diameter/2

r = 11 cm

Now, area of the circle,

$A=\pi \times (11)^2\\\\A=380.13\ cm^2$

(5) Diameter = 30 cm

Since, radius = diameter/2

r = 15 cm

Now, area of the circle,

$A=\pi \times (15)^2\\\\A=706.85\ cm^2$

(6) Diameter = 4cm

Since, radius = diameter/2

r = 2 cm

Now, area of the circle,

$A=\pi \times (2)^2\\\\A=12.56\ cm^2$

So, the area of circles 4,5 and 6 are 380.13², 706.85² and 12.56² respectively.

4 0

2 years ago

Weekly wages at a certain factory

shtirl [24]

Where is the question

6 0

3 years ago

Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0

3 years ago