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3 years ago

7

What are the steps to 43,021 -835.6?

1 answer:

Kruka [31]3 years ago

4 0

The answer says 42,185.4

Hope this helps!

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A polynomial function with a leading coefficient of 1 and multiplicity 1 for each root has these zeros:

Gala2k [10]

Answer:

1.

$x - 2 \sqrt{11}$

2.

$x + 2 \sqrt{11}$

5.

$x - 4$

Step-by-step explanation:

It is correct on Edge!

Give thanks please!

6 0

2 years ago

PLEASE HELP MEEE <br><br> :))

ivanzaharov [21]

Answer:

Question (1) is 5

Step-by-step explanation:

3x3=9

9+1=10

10

— = 5

2

5 0

2 years ago

Read 2 more answers

if the unit selling price is $2.50 and the unit cost is $1 what is the action needed to maintain the gross margin percentage whe

Nikolay [14]

The profit is originally 250% to maintain this you need to multiply the new unit cost by 2.5.
New unit cost $1.25. * 2.5 = $3.12.5

5 0

3 years ago

Obtain the general solution of<br> y ln x ln y dx + dy = 0

oksano4ka [1.4K]

Dy/dx = -y inx iny

(y in x in y) dx + dy = 0

(y in x in y ) dx = - dy

(in x )dx = - (1/ y in y) dy

you just need to integrate both sides

Hope this helps

4 0

3 years ago

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

$x^{2}+x-56=0$

$x^{2}+8x-7x-56=0$

$(x+8)-7(x+8)=0$

$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

$f(7)=2(7)^3+3(7)^2-336(7)$

$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

3 0

3 years ago