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2 years ago

Given that (7,-4) is on the graph of f(x), find the corresponding point for the function f(x+4)

Mathematics

2 answers:

enot [183]2 years ago

4 0

If x+4 is 7, then x=3. The corresponding point is (3, -4).

andrew11 [14]2 years ago

3 0

Answer:

3,-4

Step-by-step explanation:

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create a problem using radios of for your guardian/parent or friends solve. write both of your problems and Solutions in your Jo

victus00 [196]

Answer:

The following,... and uh I don't have a journal ;->

Step-by-step explanation:

You and your 10 friends each have 121 books that you will be bringing to a Best Book Barrage book exchange. Once everyone arrives you count a total of books. Assuming everyone brings the same amount of books how many people came to the exchange?

Answer: 8,228

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2 years ago

4x-3=2x+7<br><br><br> HELP PLEASEEE

Mkey [24]

Answer:

x = 5

Step-by-step explanation:

Given

4x - 3 = 2x + 7 ( subtract 2x from both sides )

2x - 3 = 7 ( add 3 to both sides )

2x = 10 ( divide both sides by 2 )

x = 5

6 0

3 years ago

Read 2 more answers

The speed of light is about 300,000 kilometers per second. The Sun is about 93 million miles from Earth. How many minutes does i

spin [16.1K]

Answer: 8.3 minutes

Step-by-step explanation:

1 mile is to 1.60934km so 93 million miles is:

= 93 * 1.609

= ‭149,637,000‬ km

Travelling at 300,000 km per second, the number of seconds it takes is:

= ‭149,637,000‬/300,000

= ‭498.79‬ secs

In minutes this is:

= ‭498.79‬/60

= 8.3 minutes

4 0

2 years ago

Consider the circle of radius 5 centered at (0, 0). Find an equation of the line tangent to the circle at the point (3, 4) in sl

Wittaler [7]

Answer:

$\displaystyle y= -\frac{3}{4} x + \frac{25}{4}$ .

Step-by-step explanation:

The equation of a circle of radius $5$ centered at $(0,0)$ is:

$x^{2} + y^{2} = 5^{2}$ .

$x^{2} + y^{2} = 25$ .

Differentiate implicitly with respect to $x$ to find the slope of tangents to this circle.

$\displaystyle \frac{d}{dx}[x^{2} + y^{2}] = \frac{d}{dx}[25]$

$\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(y^{2}) = 0$ .

Apply the power rule and the chain rule. Treat $y$ as a function of $x$ , $f(x)$ .

$\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(f(x))^{2} = 0$ .

$\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}(2f(x)\cdot f^{\prime}(x)) = 0$ .

That is:

$\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}\left(2y \cdot \frac{dy}{dx}\right) = 0$ .

Solve this equation for $\displaystyle \frac{dy}{dx}$ :

$\displaystyle \frac{dy}{dx} = -\frac{x}{y}$ .

The slope of the tangent to this circle at point $(3, 4)$ will thus equal

$\displaystyle \frac{dy}{dx} = -\frac{3}{4}$ .

Apply the slope-point of a line in a cartesian plane:

$y - y_0 = m(x - x_0)$ , where

$m$ is the gradient of this line, and
$(x_0, y_0)$ are the coordinates of a point on that line.

For the tangent line in this question:

$\displaystyle m = -\frac{3}{4}$ ,
$(x_0, y_0) = (3, 4)$ .

The equation of this tangent line will thus be:

$\displaystyle y - 4 = -\frac{3}{4} (x - 3)$ .

That simplifies to

$\displaystyle y= -\frac{3}{4} x + \frac{25}{4}$ .

3 0

2 years ago

A scuba diver diving with an 80 cubic foot tank of air, needs to resurface when his air in the

Burka [1]

At sea level atmospheric pressure is 1 bar absolute (1 standard atmosphere =101 kPa=1.013 bars). The weight of the atmosphere exerts a pressure which will support a column of water 10 m high; 10 m under water the pressure on a diver is 200 kPa. The volume of gas in an early diving bell full of air at sea level is halved at 10 m according to Boyle’s law; at 20 m pressure is 300 kPa absolute and the gas is compressed into one third the volume.

Dry air is composed of roughly 21% oxygen, 78% nitrogen, and 1% other gases. According to Dalton’s law the partial pressure of oxygen at any depth will be 21% of the total pressure exerted by the air and the partial pressure of nitrogen will be 78% of total pressure.

Gases dissolve in the liquid with which they are in contact. Nitrogen is fat soluble and at sea level we have several litres dissolved in our bodies. If the partial pressure of nitrogen is doubled (by breathing air at 10 m depth) for long enough for equilibration to take place we will contain twice as many dissolved nitrogen molecules as at sea level.

6 0

2 years ago