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3 years ago

15

You are analyzing the survey results tabulated below. The first survey question (yes/no) is whether the respondent is over 40, a

nd the second survey question (low-cal, regular) is whether the respondent prefers your low-cal butter or the regular.

1 answer:

kolbaska11 [484]3 years ago

5 0

Answer:

Step-by-step explanation:

tryin to cheat in workkeys

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Subtract: (2x2 − 7x + 5) − (−6x2 − 4x − 2)

Juli2301 [7.4K]

2x^2 - 7x + 5 - (-6x^2 - 4x - 2) =
2x^2 - 7x + 5 + 6x^2 + 4x + 2 =
8x^2 - 3x + 7 <==

8 0

3 years ago

CALC- limits<br> please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

3 years ago

A field is a rectangle with a perimeter of 1220 feet. The length is 100 feet more than the width. Find the width and length of t

rosijanka [135]

To start, let x represent the width and x+100 represent the length. Since the perimeter of a figure is the sum of all the measurements of the side which can be represented by (x+100)+(x+100)+x+x and since you know your perimeter is 1220, you can set the expression equal to 1220. This would look like this:
(x+100)+(x+100)+x+x=1220

Once you have done that, combine any like terms (combine terms with the same variables and raised to the same power together) which would simplify to this:
4x+200=1220

Now that you have your like terms simplified, subtract 200 from both sides to get 4x=1020 and finally, to solve for x, or find the width, divide both sides by 4 to get x=255.

Now that you have your width, now you must find your length as the question asks to find the dimensions of the rectangular field. To find the length, add 100 to the width, 255 since according to the information given, the length is 100 more than the width. When you add 100 to 255, you should get that your length is 355.

Now that you have your length and width, you can conclude that the dimensions of the field is 255 by 355 feet, which is your answer :)

3 0

3 years ago

If two cards are drawn from a regular shuffled deck, what is the probability that both cards will be diamonds?

Nastasia [14]

Answer: The first card is a diamond and the second card is a heart. Both cards are hearts. Let H denote the event that a heart is drawn; let D denote the event that a diamond is drawn. The first card is a diamond and the second card is a heart: The probability of drawing a diamond on the first draw

Step-by-step explanation:

6 0

2 years ago

ryzh [129]

Answer:

(1 ) Inner curved surface area of the well is 109.9 sq. meters.

(2) The cost of plastering the total curved surface area is 4396.

Step-by-step explanation:

The inner diameter = 3.5 m

Depth of the well = 10 m

Now, Diameter = 2 x Radius

⇒R = D/ 2 = 3.5/2 = 1.75

or, the inner radius of the well = 1.75 m

CURVED SURFACE AREA of cylinder = 2πr h

⇒The inner curved surface area = 2πr h = 2 ( 3.14) (1.75)(10)

= 109 sq. meters

Hence, the inner curved surface area of the well is 109.9 sq. meters.

Now, the cost of plastering the curved area is 40 per sq meters

So, the cost of total plastering total area = 109.9 x(Cost per meter sq.)

= 109.9 x (40)

= 4396

Hence, the cost of plastering the total curved surface area is 4396.

8 0

3 years ago