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3 years ago

7

The vertices of a parallelogram are at (-9, 12), (-3, 12), (-5, 10), and (–11, 10). What is the perimeter of the parallelogram?

Round your answer to the nearest whole number?

1 answer:

eduard3 years ago

7 0

let point A has coordinates (-9;12),B(-3;12),C(-5;10) and D(-11;10).Then the vector AB={6;0}, BC={2;-2}, CD={6;0}, AD={2;-2} and their and the length of the vectors

$|AB| = |CD| = \sqrt{ {6}^{2} + {0}^{2} } = \sqrt{36} = 6 \\ |BC| = |AD| = \sqrt{ {2}^{2} + {( - 2)}^{2} } = \sqrt{8}$

$p = 6 + 6 + \sqrt{8} + \sqrt{8} = \\ = 12 + 2 \sqrt{8} = 17.66\approx18$

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Please help with at least one question.

Cloud [144]

Okay, first off negative is well negative and since there is no other negatives, -5/7 is least (making it go first). Second, what is smaller 3/5 or 2/9, remember that the bigger the denominator the smaller the piece.

[/] [/] [/] [] [] = 3/5

[/] [/] [] [] [] [] [] [] [] 2/9

At the end it should be -5/7, 2/9 , 3/5. This method can go the same for problems like these...Hope it helps!

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3 years ago

Which ski run would probably give you the greatest speed down a hill when you are skiing: one with a slope of 1/8, 1/4, 1, or 2?

saul85 [17]

Answer:

2

Step-by-step explanation:

For Every x value the y value increases by 2

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2 years ago

A box in the shape of a roctangular prism has a longth of 3 inches, a width of 2, inches,

tatyana61 [14]

I need help Over the next two days, Benton Employment Agency is interviewing clients who
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3 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$

$f'(x)=12x^2+26x-12$

$f''(x)=24x+26$

So if f''(a) is positive then we have a minimum at x=a.

If f''(a) is negative then we have a maximum at x=a.

Rounding to nearest tenths here: x=-2.6 and x=.4

Let's see what f'' gives us at both of these x's.

$24(-2.6)+25$

$-37.5$

So we have a maximum at x=-2.6.

$24(.4)+25$

$9.6+25$

$34.6$

So we have a minimum at x=.4.

Now let's find the corresponding y-value for our relative minimum point since that would complete your question.

We are going to use the equation that relates x and y.

I'm going to use 0.3909838 instead of .4 just so we can be closer to the correct y value.

$y=4(0.3909838)^3+13(0.3909838)^2-12(0.3909838)-56$

I'm shoving this into a calculator:

$y=-58.4654411$

So the approximate relative minimum is (0.4,-58.5).

If you graph $y=4x^3+13x^2-12x-56$ you should see the graph taking a dip at this point.

3 0

3 years ago

A rectangular lawn has an area of 140 square yards. Its width is 6 yards less

Katarina [22]

Answer:

length : 10 yard

width: 14 yard

Step-by-step explanation:

Let l be the length of the yard.

Since the width is 6 yards less than twice its length, we can say the width is 2l-6.

With width x length=area,

l (2l -6) = 140

2l^2 - 6l = 140

2l^2 - 6l - 140 = 0

Using the quadratic formula,

we can determine l = 10 or -7

since the length should not be negative,

the length is 10 yard.

Now substitute l=10 into 2l-6.

width = 2(10)-6

= 14 yard

6 0

3 years ago