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3 years ago

A line passes through the point (3,6) and has a slope of 5?

Mathematics

2 answers:

lbvjy [14]3 years ago

8 0

Answer:

y = 5x - 9

Step-by-step explanation:

To write an equation in slope-intercept form, two values are needed: the y-intercept and the slope. Since we already have the slope, to find the equation of this line, we need to find the y-intercept.

To find the y-intercept, plug the point into the equation for slope-intercept, which is y = mx + b. m = slope and b = y-intercept.

6 = 5(3) + b

Now, isolate b.

Evaluating 5*3, we get 6 = 15 + b.

Subtracting 15 from both sides, we get -9 = b.

Now, plug b and m into the equation for slope-intercept form to get your final answer:

y = 5x - 9

hammer [34]3 years ago

6 0

Answer:

y = 5x - 9

Step-by-step explanation:

y - y₁ = m(x - x₁), where (x₁, y₁) = (3, 6)

y - 6 = 5(x - 3)

y = 5x - 15 + 6

y = 5x - 9

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3 years ago

Choose the abbreviation of the postulate or theorem that supports the conclusion that WAS NOT. Given: WS = NT, AS = OT, S = T.

koban [17]

Answer: The abbreviation that supports the conclusion is SAS

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So the postulate that supports is SAS

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3 years ago

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lozanna [386]

Answer:

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2 years ago

Read 2 more answers

2. The population of Small Town, Alabama was 13,128 in 1990 but the town is growing at a rate

belka [17]

Answer i've done the math multiple times and everytime i get the answer 21,950. i don't think i'm doing the math wrong, maybe try asking the teacher or whoever is the one grading it cause it could be a typo or something like that.

Step-by-step explanation:

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3 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 $\frac{dP}{dt}=3.68$ .

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3 years ago