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Dahasolnce [82]
3 years ago
10

PLEASEEEE HELPPP!!

Mathematics
1 answer:
viktelen [127]3 years ago
5 0

Answer:distributive

Step-by-step explanation:

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I need help with this one
Gwar [14]
Factors always multiply each other.

The only option that multiplies something is (y^3 + 3)
So the answer is B.
8 0
3 years ago
Emma took a taxi from the airport to her house.
Natali5045456 [20]
Hi there! The answer is 10 miles.

Let the distance be represented by b.
The cost if the ride will therefore be
$5 + ( $3.70 × d )

We end up with the following equation
5 + 3.7d = 42

Subtract 5 from both sides.
3.7d = 37

Divide both sides by 10.
d =  \frac{37}{3.7}  = 10

Therefore, Emma had a taxi ride of 10 miles.
8 0
3 years ago
Read 2 more answers
What is 16 – x at x = 5?
Sunny_sXe [5.5K]

Answer:

11

Step-by-step explanation:

16-x

substitute with 5

16-5

11

8 0
3 years ago
Devin is investigating the relationship between the temperature in degrees Fahrenheit and the number of gallons sold at a lemona
Temka [501]

Answer:

d. A good line of best fit is y= 0.36x -20.15 because this is what the regression calculator returns if all the data except the point (86, 2) are entered. :)

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).
gtnhenbr [62]

The slope of the tangent line to the curve at (8, 2) is given by the derivative \frac{dy}{dx} at that point. By the chain rule,

\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}

Differentiate the given parametric equations with respect to t :

x = 4t \implies \dfrac{dx}{dt} = 4

y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}

Then

\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}

We have x=8 and y=2 when t=2, so the slope at the given point is \frac{dy}{dx} = -\frac14.

The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}

Alternatively, we can eliminate the parameter and express y explicitly in terms of x :

x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x

Then the slope of the tangent line is

\dfrac{dy}{dx} = -\dfrac{16}{x^2}

At x = 8, the slope is again -\frac{16}{64}=-\frac14, so the normal has slope +4, and so on.

5 0
2 years ago
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