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2 years ago

14

Please help me i don't know how to do this no more

1 answer:

aalyn [17]2 years ago

5 0

Answer:

z/3 = 3/2 so 2z = 9 z = 4.5 cuz triangles are proportional

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Lola takes the train from Paris to Nice. The distance between the two cities is about 900,000 meters. If the train travels at a

Elanso [62]

Nice??? there is a city called nice? anyways, your answer is 5000. hope this helps :)

4 0

3 years ago

Read 2 more answers

Please solve for me.

blsea [12.9K]

Answer:

570

Step-by-step explanation:

3 0

3 years ago

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A function is shown.<br><br> f(x)=−3⋅4x<br> What is the value of x when f(x) = −48?

galben [10]

Answer:

576

Step-by-step explanation:

Whenever you are solving these types of problems, you just put the value given in place of x. For this equation you would have done f(48)=-3·4(48)

6 0

3 years ago

A student takes a multiple choice test with 40 questions. The probability that a student answers a given question correctly is 0

LenKa [72]

Answer:

Step-by-step explanation:

X no of questions student answers is binomial with n =40 and p =0.5

If approximated to normal, X is Normal with

mean = np = 20 and variance = npq = 10

P(X>N) >0.10

We use std normal distribution table to get z value first then convert to x value

$z>-1.28$

So $x>20-1.28(10) = 20-12.8 = 7.2$

This is with continuity correction.

Hence without continuity correction this equals 7.2-0.5 = 6.7

x>6.7

n = 7

6 0

3 years ago

(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