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3 years ago

Write the equation of the line. Slope = -4, passing through (- 1, 5)

Mathematics

1 answer:

Vladimir79 [104]3 years ago

3 0

Answer:

y=-4x+1

Step-by-step explanation:

You want to find the equation for a line that passes through the point (-1,5) and has a slope of -4.

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

To start, you know what m is; it's just the slope, which you said was -4. So you can right away fill in the equation for a line somewhat to read:

y=-4x+b.

Now, what about b, the y-intercept?

To find b, think about what your (x,y) point means:

(-1,5). When x of the line is -1, y of the line must be 5.

Because you said the line passes through this point, right?

Now, look at our line's equation so far: . b is what we want, the -4 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the the point (-1,5).

So, why not plug in for x the number -1 and for y the number 5? This will allow us to solve for b for the particular line that passes through the point you gave!.

(-1,5). y=mx+b or 5=-4 × -1+b, or solving for b: b=5-(-4)(-1). b=1.

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3 years ago

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The ages of the 32 recruits in police academy are normally distributed with a mean of 27 with a standard deviation of 2. What is

aksik [14]

Since it's a sample size, we use x, as the mean and s= standard deviation
_
Given: x =27, s = 2 and total sample = 32.
Probability that a recruit is at least 31 years old?

Let's calculate the Z score :
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X - x 31 - 27
Z = ----------- → Z = -------- → Z = 2 (now look up in the Z score table,
s 2

Area (Z=2) =0.9772 , but Z ≥ 2 (at least) , then :
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3 years ago

Is 5.875 bigger than 5.8

tamaranim1 [39]

Yes it is. Hope this helps

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3 years ago

Use calculus to find the absolute maximum and minimum values of the function. (round all answers to three decimal places.) f(x)

Allisa [31]

Part A:

Given the function $f(x)=x+2\cos(x)$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6}$

Using the second derivative test,

$f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732$

Since the second derivative gives a negative number, the given function has a maximum point at $x=\frac{\pi}{6}$ .

And the maximum point is given by:

$f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}$

i.e. $\left(\frac{\pi}{6},\ 2.256\right)$

Part B:

Given the function $f(x)=e^{-x}-e^{-2x}$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931$

Using the second derivative test,

$f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5$

Since the second derivative gives a negative number, the given function has a maximum point at $x=0.6931$ .

And the maximum point is given by:

$f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}$

i.e. (0.693, 0.25)

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3 years ago

Ann took a taxi home from the airport. The taxi fare was $2.10 per mile, and she gave the driver a tip of $5. Ann paid a total o

Dmitry_Shevchenko [17]

$2,10 / mile
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x - <span>the distance in miles
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x=21$

<span>Distance between airport and Ann's home is 21 miles.
</span>

Maja

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3 years ago

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