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3 years ago

Graph g(x) = –2x – 3

Mathematics

1 answer:

Luba_88 [7]3 years ago

4 0

Answer:

The slope would be -2 with a y-intercept of (0, -3).

I hope this helped to answer your question.

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What is the product?<br> (-3s+2t)(4s-t)

Marizza181 [45]

Answer:

12s² + 11st - 2t²

Step-by-step explanation:

Each term in the second factor is multiplied by each term in the first factor, that is

- 3s(4s - t) + 2t(4s - t) ← distribute both parenthesis

= - 12s³ + 3st + 8st - 2t² ← collect like terms

= - 12s² + 11st - 2t²

5 0

4 years ago

Read 2 more answers

Analyze the graph of a cannonball launched from the ground with an initial velocity of 144 feet per second. The cannonball reach

MrRissso [65]

We can find that the formula is:
$h(t)=-16t^2 + 144t$

We are told that the cannonball reaches its maximum height at t = 4.5 seconds. To find the maximum height, let's just plug this into the formula:
$h(4.5) = -16(4.5)^2 + 144(4.5) = 324$

The height is 324 feet. We're done!

7 0

3 years ago

Read 2 more answers

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

A radio is being discounted 30%. If the sale price of the radio is $29.82, what was the original price of the radio?

meriva

Answer: $42.60

Step-by-step explanation:

Original price: p

Discount: 30%

Sale price: 29.82

p x 0.7 = 29.82

p = 29.82/0.7 = 42.60

3 0

3 years ago

NEED HELP WILL MARK BRAINIEST!

Illusion [34]

Answer:

A) 23+5=28

B) 23-5=18

i hope it helped

7 0

3 years ago