What is 0.9 1/ 10th of

Mathematics

2 answers:

kozerog [31]3 years ago

8 0

Maybe 0.09
i dont know because you worded it weird

Vitek1552 [10]3 years ago

7 0

1.0 I think I'm young I'm not to smart

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How can 563*10 to the 3rd power be scientific notation.(Use the photo above please)

zaharov [31]

563*(10*10*10) that's you scientific notation

6 0

3 years ago

rewona [7]

X = 5/3 and m∠SPY = 8 1/3°.

Since PQ bisects the angle, the two angles formed by the bisector are equal:

11/2x - 5 = 4x - 5/2

We will first multiply everything by 2 to eliminate the fractions:
(11/2x)*2 - 5*2 = 4x*2 - (5/2)*2
11x - 10 = 8x - 5

Subtract 8x from each side:
11x - 10 - 8x = 8x - 5 - 8x
3x - 10 = -5

Add 10 to both sides:
3x - 10 + 10 = -5 + 10
3x = 5

Divide both sides by 3:
3x/3 = 5/3
x = 5/3

Now we will plug this in for x in both smaller angles and add them together to find the measure of ∠SPY:

11/2(5/3) - 5 + 4(5/3) - 5/2
55/6 - 5 + 20/3 - 5/2

We will rewrite everything using a denominator of 6:
55/6 - 30/6 + 40/6 - 15/6 = (55-30+40-15)/6 = 50/6 = 8 2/6 = 8 1/3°

3 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:

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