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2 years ago

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A line passes through the point (2,-5) and is perpendicular to the line with the

1 answer:

zloy xaker [14]2 years ago

8 0

Answer:

D) y = -5x + 5

Step-by-step explanation:

If two lines are perpendicular to each other, they have opposite slopes.

The first line is y = 1/5x + 5. Its slope is 1/5. A line parallel/perpendicular to this one will have a slope of -5.

Plug this value (#) into your standard point-slope equation of y = mx + b.

y = -5x + b

To find b, we want to plug in a value that we know is on this line: in this case, it is (2, -5). Plug in the x and y values into the x and y of the standard equation.

-5 = -5(2) + b

To find b, multiply the slope and the input of x (2)

-5 = -10 + b

Now, add 10 to both sides to isolate b.

5 = b

Plug this into your standard equation.

y = -5x + 5

This equation is parallel/perpendicular to your given equation (y = 1/5x + 5) and contains point (2, -5)

Hope this helps!

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Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = e−4x, [0, 2] Yes, it does not m

Zigmanuir [339]

Answer:

(a) Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem.

(b) $c =0.51995$

Step-by-step explanation:

Given

$f(x) = e^{-4x};\ [0,2]$

Solving (a); Does the function satisfy M.V.T on the given interval

We have:

$f(x) = e^{-4x};\ [0,2]$

The above function is an exponential function, and it is differentiable and continuous everywhere

Solving (b): The value of c

To do this, we use:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

In this case:

$[a,b] = [0,2]$

So, we have:

$f'(c) = \frac{f(2) - f(0)}{2 - 0}$

$f'(c) = \frac{f(2) - f(0)}{2}$

Calculate f(2) and f(0)

$f(x) = e^{-4x}$

So:

$f(2) = e^{-4*2} = e^{-8} = 0.00033546262$

$f(0) = e^{-4*0} = e^{0} = 1$

This gives:

$f'(c) = \frac{0.00033546262 - 1}{2}$

$f'(c) = \frac{-0.99966453738}{2}$

$f'(c) = -0.4998$

Note that:

$f'(x) = (e^{-4x})'$

$f'(x) = -4e^{-4x}$

This implies that:

$f'(c) = -4e^{-4c}$

So, we have:

$f'(c) = -0.4998$

$-4e^{-4c} =-0.4998$

Divide both sides by -4

$e^{-4c} =\frac{-0.4998}{-4}$

$e^{-4c} =0.12495$

Take natural logarithm of both sides

$\ln(e^{-4c}) =\ln(0.12495)$

$\ln(e^{-4c}) =-2.0798$

Apply law of natural logarithm

$\ln(e^{ax}) =ax$

So:

$-4c =-2.0798$

Solve for c

$c =\frac{-2.0798}{-4}$

$c =0.51995$

3 0

2 years ago

Find the surface area of a wooden box full of shape is of a cube and if the age of the box is 4 cm

Sav [38]

All you have to do is multply 4 by all the sides of the cube

4 0

2 years ago

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What is 10(b + 4c) expanded? (using the distributive property)

jeka94

Answer:

10b+40c

Step-by-step explanation:

10(b + 4c) you have to multiply 10 by each member

10b+40c

8 0

2 years ago

I need help on my trig can someone help me on the question please and thank you

IRINA_888 [86]

Answer:

210 degrees

Step-by-step explanation:

Convert the problem from radians to degrees using the ratio 180/ and it will give you 210.

6 0

2 years ago

Sanjay graphs a quadratic function that has x-intercepts of –3 and 7. Which functions could he have graphed?

kykrilka [37]

we will select each options and find zeros

(a)

$g(x)=x^2-4x-21$

for finding x-intercept , we can set g(x)=0

and then we can solve for x

$g(x)=x^2-4x-21=0$

now, we can factor it

$(x-7)(x+3)=0$

we get

$x=7,x=-3$

so, this is TRUE

(b)

$g(x)=(x-3)(x+7)$

we can set it to 0

and then we can solve for x

$g(x)=(x-3)(x+7)=0$

we get

$x=3,x=-7$

so, this is FALSE

(c)

$g(x)=3x^2-12x-63$

we can set it to 0

and then we can solve for x

$g(x)=3x^2-12x-63 =0$

$3(x^2-4x-21) =0$

$3(x-7)(x+3) =0$

$x=-3,x=7$

so, this is TRUE

(d)

$g(x)=-(x+3)(x-7)$

now, we can set it to 0

and then we can solve for x

$g(x)=-(x+3)(x-7)=0$

$x=-3,x=7$

so, this is TRUE

(e)

we have

$g(x)=x^2+4x-21$

now, we can set it to 0

and then we can solve for x

$g(x)=x^2+4x-21=0$

$(x+7)(x-3)=0$

$x=-7,x=3$

so, this is FALSE

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2 years ago

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