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Tasya [4]
3 years ago
7

248.92 x 10 to 4th power

Mathematics
2 answers:
Karo-lina-s [1.5K]3 years ago
7 0

Answer:

2489200

Step-by-step explanation:

First you have 248.92. You also have 10^4

Your equation should look like 248.92 * 10^4

Calculate using a calculator and you should arrive at the answer 2,489,200.

faltersainse [42]3 years ago
4 0

38,382,608,135,847.8

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8-4x=4-3(2x+6) ??? Someone help
Maurinko [17]
<span>       8 - 4x=4 - 3(2x+6)
<=> 8 - 4x = 4 - 6x - 18
<=> 8 - 4 + 18 = 4x - 6x
<=> 22 = -2x
x = -11</span>
7 0
3 years ago
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3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k
Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt

Evaluate the integral to solve for y :

\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt

\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x

\displaystyle y = x^3+2x^2+kx - 2

Use the other known value, f(2) = 18, to solve for k :

18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}

Then the curve C has equation

\boxed{y = x^3 + 2x^2 + 2x - 2}

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2

The slope of the given tangent line y=x-2 is 1. Solve for a :

3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1

So, the point of contact between the tangent line and C is (-1, -3).

7 0
2 years ago
During which one-day interval did she work the most hours?
Karo-lina-s [1.5K]
Um well lets see there is no question so the day that ends in y
5 0
3 years ago
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A converging lens of focal length 5 cm is placed at a distance of 20
lesya [120]

The object should be placed 20/cm from the lens so as to form its real image on the screen

<h3>Mirror equation</h3>

The mirror equation is given according to the expression below;

1/f = 1/u + 1/v

where

f is the focal length

u is the object distance

v is the image distance

Given the following parameters

f = 5cm

v = 20cm

Required

object distance u

Substitute

1/5 = 1/u +1/20

1/u = 1/5-1/20
1/u = 3/20

u = 20/3 cm

Hence the object should be placed 20/cm from the lens so as to form its real image on the screen

Learn more on mirror equation here; brainly.com/question/27924393

#SPJ1

4 0
2 years ago
Due in 1 hour, plz help
snow_lady [41]

Answer:

1. 12/1= rate of change

Step-by-step explanation:

1.

1 mile= 30 seconds

(1,30)

6 miles= 1 1/2 minutes

6 miles=90seconds

(6,90)

Formula: y2-y1/x2-x1

90-30/6-1

60/5

12/1= rate of change

hope its right

3 0
3 years ago
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