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2 years ago

What are the coordinates of the midpoint of EF?

Mathematics

1 answer:

BabaBlast [244]2 years ago

7 0

The answer would be d. (1,3.5)
you would use the midpoint formula that is attached below.

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Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

3 years ago

The absolute value function, f(x) = –|x| – 3, is shown. What is the range of the function? all real numbers all real numbers les

Harman [31]

Answer:

all real numbers less than or equal to –3

Step-by-step explanation:

Look at the y-values that the graph shows. There are none greater than -3. Any value of -3 or less is possible.

5 0

2 years ago

Read 2 more answers

(WHO EVER HELPS ME GETS BRAINLIEST AND EXTRA POINTSSSSS :D )

Romashka [77]

Answer:d

D. 9 and 46/180

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Change to log form. <br> 7e^(-x)=0.01

sladkih [1.3K]

(0.00142857)=-x is your answer

3 0

2 years ago

A rectangle is twice as long as it is wide. If its length and width are both

puteri [66]

Answer:

a rectangle is twice as long as it is wide . if both its dimensions are increased 4 m , its area is increaed by 88 m squared make a sketch and find its original dimensions of the original rectangle

Step-by-step explanation:

Let l = the original length of the original rectangle

Let w = the original width of the original rectangle

From the description of the problem, we can construct the following two equations

l=2*w (Equation #1)

(l+4)*(w+4)=l*w+88 (Equation #2)

Substitute equation #1 into equation #2

(2w+4)*(w+4)=(2w*w)+88

2w^2+4w+8w+16=2w^2+88

collect like terms on the same side of the equation

2w^2+2w^2 +12w+16-88=0

4w^2+12w-72=0

Since 4 is afactor of each term, divide both sides of the equation by 4

w^2+3w-18=0

The quadratic equation can be factored into (w+6)*(w-3)=0

Therefore w=-6 or w=3

w=-6 can be rejected because the length of a rectangle can't be negative so

w=3 and from equation #1 l=2*w=2*3=6

I hope that this helps. The difficult part of the problem probably was to construct equation #1 and to factor the equation after performing all of the arithmetic operations.

5 0

2 years ago