<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20x%20%5Cfrac%7Bx%5E%7B2%7D-x-2%7D%7Bx-2%7D" id="TexFormula1" title="\lim

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Rudik [331]

3 years ago

to x \frac{x^{2}-x-2}{x-2}" alt="\lim_{x\to x \frac{x^{2}-x-2}{x-2}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

FromTheMoon [43]3 years ago

8 0

Answer:

Step-by-step explanation:

where is the question there is just says lim ?

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Find the value of 11 +22+33 + ... + 1100.

ella [17]

Answer:

sum of terms: 55550

Explanation:

$\sf S_n = \dfrac{n}{2} (u_{start}+u_{end}})$

Here the number of terms:

1100 ÷ 11 = 100

Using the Formula:

$\sf \dfrac{100}{2} (11+1100)$

$\sf 55550$

5 0

2 years ago

Does anyone understand this?. I also have to show my work.

snow_lady [41]

Answer:

Ans A). The graph is shown.

Ans B). 18.3333 C temperature when F is 65 temperature

Ans C). 32 F when the line crosses the horizontal axis

Ans D). Slope of line C= $\frac{5}{9}(F -32)$ is $\frac{5}{9}$

Step-by-step explanation:

Given equation is C= $\frac{5}{9}(F -32)$

Ans A).

For the table,

Take the four value of F as 32,41,50,59.

For F = 32.

The value of C is

C= $\frac{5}{9}(F -32)$

C= $\frac{5}{9}(32 -32)$

C=0.

For F = 41.

The value of C is

C= $\frac{5}{9}(F -32)$

C= $\frac{5}{9}(41 -32)$

C=05

For F = 50.

The value of C is

C= $\frac{5}{9}(F -32)$

C= $\frac{5}{9}(50 -32)$

C=10

For F = 59.

The value of C is

C= $\frac{5}{9}(F -32)$

C= $\frac{5}{9}(32 -32)$

C=15

<em>Note: The figure shows a graph of given equation with points.</em>

Ans B). Estimate temperature in C when the temperature in F is 65

For F = 65.

The value of C is

C= $\frac{5}{9}(F -32)$

C= $\frac{5}{9}(32 -32)$

Ans C). At what temperature, graph lien cross the horizontal axis

When the line crosses the horizontal axis, C=0

Therefore,

C= $\frac{5}{9}(F -32)$

0= $\frac{5}{9}(F -32)$

0= $(F -32)$

F=32 Temperature.

Ans D). Slope of the line C= $\frac{5}{9}(F -32)$

The slope of line is given by s= $\frac{Y1-Y2}{X1-X2}$

Take points from the table of answer A.

let (32,0) and (41,5) using for slope.

s= $\frac{Y1-Y2}{X1-X2}$

s= $\frac{0-5}{32-41}$

s= $\frac{5}{9}$

Slope of line C= $\frac{5}{9}(F -32)$ is $\frac{5}{9}$

7 0

3 years ago

The probability that a customer of a network operator has a problem about you needing technical staff's help in a month is 0.01.

snow_tiger [21]

Answer:

(a) average calls = 5

(b) probability that there is exactly one call in 6 consecutive monts = 0.038

Step-by-step explanation:

Let event of a customer requiring help in a particular month = H

and event of a customer not requiring help in a particular month = ~H

Given

p= 0.01, therefore

Number of households, n = 500.

Binomial distribution:

x = number of households requiring help in a particular month

P(x,n,p) = C(x,n)*p^x*(1-p)^(n-x)

where

C(x,n) = n!/(x!(n-x)!) is the the number of combinations of x objects out of n

(a) Average number of households requiring help = np = 500*0.01 = 5

(b)

Probability that there are no calls requiring help in a particular month

P(0), q= C(0,n)*p^0(1-p)^(n-0)

= 1*1*0.99^500

= 0.006570483

Applying binomial probability over six months,

q = 0.006570483

n = 6

x = 1

P(x,n,q)

= C(x,n)*q^x*(1-q)^(n-x)

= 6!/(1!*5!) * 0.006570483^1 * (1-0.006570483)^5

= 0.038145

Therefore the probability that in 6 consecutive months there is exactly one month that no customer has called for help = 0.038

3 0

3 years ago

8.) Plot its cost function

ss7ja [257]

The attached shows the two plots. The break-even points are near
x = 21
x = 256

5 0

3 years ago

Read 2 more answers

PLEASE I need help GeoMETRY

9966 [12]

Answer:

2. 2.3

3. 37.2 m

Step-by-step explanation:

2. tan theta = opposite/ adjacent

tan 25 = x/5

Multiply each side by 5

5 tan 25 = x/5 *5

5 tan 25 = x

2.331538291

To the nearest tenth

2.3 =x

3. We want to find the height or y

sin theta = opposite/ hypotenuse

sin 60 = y /43

Multiply each side by 43

43 sin 60 = y/43 * 43

43 sin 60 = y

37.23909236 = y

To the nearest tenth

37.2 =y

The light pole is 37.2 meters tall

5 0

3 years ago