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2 years ago

5.1x10^-1 as an ordinary number?

Mathematics

2 answers:

scZoUnD [109]2 years ago

8 0

=0.51=51/100............

REY [17]2 years ago

7 0

10^-1 = 0.1

so its 5.1 * 0.1 = 0.51 Answer

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2. The restaurant bill you and your friend received was not itemized. You ordered 2 drink refills and the egg breakfast, for a t

fiasKO [112]

( a ) The system of equations:
2 r + b = 8.40
3 r + b = 9.35
( b ) Graph is in the attachment.
( c ) Each item costs:
b = $6.50, r = $0.95
We can prove it: 2 * 0.95 + 6.50 = 8.40
3* 0.95 + 6.50 = 9.35

Download docx

7 0

2 years ago

What is the following product? RootIndex 3 StartRoot 16 x Superscript 7 Baseline EndRoot times RootIndex 3 StartRoot 12 x Supers

Nadya [2.5K]

Answer:

$4x^5\sqrt[3]{3x}$

Step-by-step explanation:

I'm not sure I understand the problem, but I think it's this:

$\sqrt[3]{16x^7} \times \sqrt[3]{12x^9} =$

$= \sqrt[3]{16 \times 12 \times x^{16}}$

$= \sqrt[3]{192 \times x^{15} \times x}$

$= \sqrt[3]{64 \times 3 \times (x^5)^3 \times x}$

$= \sqrt[3]{4^3 \times 3 \times (x^5)^3 \times x}$

$= 4x^5\sqrt[3]{3x}$

8 0

1 year ago

A company produces and sells 211,600 boxes of t-shirts each year. Each production run has a fixed cost of $400 and an additional

Illusion [34]

Answer: $x=46$

Step-by-step explanation:

Given

Company Produces and sells 211600 boxes of T-shirt each year

Fixed cost =$ 400

Additional cost = $ 3 per box

storage cost =$ 2

let x be the no of production run

therefore

Holding cost per year $=holding\ cost\times average\ holding\ items=2\times \frac{211600}{2x}=\frac{211600}{x}$

$Yearly\ ordering\ cost=cost\ during\ each\ order\times number\ of\ order\ Placed\ per\ year$

yearly ordering cost $=400x+3\times \frac{211600}{x}$

Total cost C(x) $=\frac{211600}{x}+400x+3\times \frac{211600}{x}$

differentiate C(x) w.r.t to x we get

$\frac{\mathrm{d} C(x)}{\mathrm{d} x}=400-\frac{3\times 211600}{x^2}-\frac{1}{211600}$

Put $\frac{\mathrm{d} C(x)}{\mathrm{d} x}=0$ to get max/min value

$400-\frac{211600}{x^2}-\frac{3\times 211600}{x^2}=0$

$x^2=\frac{211600}{100}$

$x=\sqrt{2116}$

$x=46$

therefore 46 runs must be performed

5 0

1 year ago

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

3 0

1 year ago

Which is the graph of the system x + 3y > –3 and y < One-halfx + 1?

fenix001 [56]

Answer: THE GRAPH IS ATTACHED.

Step-by-step explanation:

We know that the lines are:

$x + 3y=-3$

$y = \frac{1}{2} x + 1$

Solving for "y" from the first line, we get:

$3y=-x-3\\\\y=-\frac{x}{3}-1$

In order to graph them, we can find the x-intercepts and the y-intercepts.

For the line $x + 3y=-3$ the x-intercepts is:

$0=-\frac{x}{3}-1\\\\(1)(-3)=x\\\\x=-3$

And the y-intercept is:

$y=-\frac{0}{3}-1\\\\y=-1$

For the line $y=\frac{1}{2} x + 1$ the x-intercepts is:

$0=\frac{1}{2} x + 1\\\\-1(2)=x\\\\x=-2$

And the y-intercept is:

$y=\frac{1}{2} (0)+ 1\\\\y=1$

Now we can graph both lines, as you can observe in the image attached (The symbols and $>$ indicates that the lines must be dashed).

By definition, the solution is the intersection region of all the solutions in the system of inequalities.

6 0

1 year ago

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