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3 years ago

The expression 3r - 6 can be written as _____. 3(r - 2) 3(r - 6) 3(3r - 6).

Mathematics

1 answer:

BaLLatris [955]3 years ago

5 0

Answer:

The expression 3r - 6 can be written as 3(r - 2)

Step-by-step explanation:

We know that 3r-6=3(r-2) because 3*r is the same as 3r and we know that 3*-2 is -6 hence making them equal. And to check it we know that 3r-6 is equal to 3r-6.

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A concert cleanup crew has 6 workers and one cleanup kit.

myrzilka [38]

Answer:

Option (D)

Step-by-step explanation:

At every 6 crew workers number of clean up kits required = 1

Therefore, for 1 crew worker number of kits required = $\frac{1}{6}$

And number of kits required for w workers = $\frac{w}{6}$ = w ÷ 6

If w = 54

Number of kits required = $\frac{54}{6}=9$

If w = 60

Number of kits required = $\frac{60}{6}=10$

If w = 66

Number of kits required = $\frac{66}{6}=11$

Therefore, table given in Option (D) is the correct table.

6 0

4 years ago

Round 78482 to the nearest ten thousand

nekit [7.7K]

Answer:

80000

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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.

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