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3 years ago

Find k given that these three points are collinear. A(0, -2), B(2, 0), and C(5, k)

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

7 0

Answer:

k = 3.

Step-by-step explanation:

If they are collinear the slope of AB = the slope of BC, so :-

(0- (-2)) / (2 - 0) = (k - 0) / (5 - 2)

2/2 = k/3

1 = k/3

k = 3.

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D which is the last one 6z^3+z^2-9z+10

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3 years ago

The function is given by the formula f(x)=3x+2.

Naily [24]

Answer:

(a) f(7) = 23

(b) x = 13/3

Step-by-step explanation:

Step 1: Define

f(x) = 3x + 2

(a) f(7) is x = 7

(b) f(x) = 15

Step 2: Substitute and Evaluate

(a) f(7) = 3(7) + 2

f(7) = 21 + 2

f(7) = 23

(b) 15 = 3x + 2

13 = 3x

x = 13/3

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4 years ago

Consider the enlargement of the pentagon. A small pentagon has a bottom side length of x centimeters and left side length of 7 c

LuckyWell [14K]

When the pentagon is drawn to scale then its value of x will be x=3.3 cm

<h3>What is the pentagon?</h3>

The pentagon is defined as the shape having five sides connected together.

Now it is given in the question that

For small pentagone

Bottom =x

left =7 cm

For larger pentagone

Bottom=7cm

left=15cm

Now since the pentagon is actually dilated from its size to the new bigger size. It means that the equivalent ratio of both the pentagons will be equal.

For smaller pentagon

$\rm Equivalent \ Ratio = \dfrac{Bottom}{Left} =\dfrac{x}{7}$

For bigger pentagon

$\rm Equivalent \ Ratio = \dfrac{Bottom}{Left} =\dfrac{7}{15}$

Since both, the ratio is the same

$\dfrac{x}{7} =\dfrac{7}{15}$

$x=\dfrac{49}{15} =3.3\ cm$

Thus the pentagon is drawn to scale than its value of x will be x=3.3 cm

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2 years ago

((Use the graph shown below to fill in the blank.))

Valentin [98]

The point (4,y) has a known x coordinate of x = 4. The y coordinate is unknown right now so we'll just call it y.

Draw a vertical line (see side note below) through 4 on the x axis. I've done so in red (see attached image). The red line crosses the graph at the point (4,1) so this tells us that y = 1

Answer: y = 1

Side Note: you don't have to draw a vertical red line but it's handy to see how it works out. After you get used to these types of problems, you can visually be able to see the answer without these extra lines.

6 0

4 years ago

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

4 years ago