What is the name of the solid figure you will make if you fold the net shown?

liq [111]

Answer:

It is 5/5 or 1/1 as it goes up 5 and to the left 5!!!

5 0

2 years ago

A small accounting firm has 4 accountants who each earn a different salary between 50,000 and 60,000 and a 5th accountant who wo

Nimfa-mama [501]

Answer: B

Step-by-step explanation:

Both the mean and the median will decrease, but the mean will decrease by more than the median.

4 0

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

Match the system of equations on the left with the number of solutions on the right

erastova [34]

Answer:

top to bottom, the answers are b, c, a

Step-by-step explanation:

One way to find the solution to a system of equations is to substitute values in. For the first one,

y=2x+3

y=2x+5,

we can substitute 2x+3 =y into the second equation to get

y=2x+5

2x+3 = 2x+5

subtract 2x from both sides

3 = 5

As 3 is not equal to 5, this is never equal and therefore has no solution

For the second one,

y= 2x+7

y = (-2/3)x + 10

We can plug y=2x+7 into the second equation to get

2x + 7 = y = (-2/3)x + 10

2x + 7 = (-2/3)x + 10

add (2/3)x to both sides to make all x values on one side

2x + (2/3)x + 7 = 10

subtract 7 from both sides to make only x values on one side and only constants on the other

2x + (2/3)x = 3

(6/3)x + (2/3)x = 3

(8/3)x = 3

multiply both sides by 3 to remove a denominator

8x = 9

divide both sides by 8 to isolate x

x=9/8

There is only one value for when the equations are equal, so this has one solution

For the third one

y = x-5

2y = 2x - 10

Plug x-5 = y into the second equation

2 * y= 2*(x-5)

2 * (x-5) = 2x - 10

2x-10 = 2x-10

add 10 to both sides

2x=2x

As 2x is always equal to 2x, no matter what x is, there are infinitely many solutions for this system

6 0

3 years ago

Absolute value of -|-12|-|4-8|

bearhunter [10]

Absolute value makes everything inside the sign positive. Therefore,
-|-12|=-12
|4-8|=|-4|=4
Therefore, -12-4=-16, But absolute value of that would be 16

6 0

2 years ago

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