Is w = 8 a solution to the equation 5 + w = 58? Explain

Answer:

[tex]\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}[/tex]

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Constant]:                                                                                             [tex]\displaystyle \lim_{x \to c} b = b[/tex]

Limit Rule [Variable Direct Substitution]:                                                             [tex]\displaystyle \lim_{x \to c} x = c[/tex]

Limit Property [Addition/Subtraction]:                                                                   [tex]\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)[/tex]

L'Hopital's Rule

Differentiation

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Basic Power Rule:

Step-by-step explanation:

We are given the following limit:

[tex]\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}[/tex]

Let's substitute in x = -2 using the limit rule:

[tex]\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}[/tex]

Evaluating this, we arrive at an indeterminate form:

[tex]\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}[/tex]

Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:

[tex]\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}[/tex]

Substitute in x = -2 using the limit rule:

[tex]\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{3(-2)^2}{4(-2)^3}[/tex]

Evaluating this, we get:

[tex]\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}[/tex]

And we have our answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit:  Limits

Answer:

10°

Step-by-step explanation:

360/36 = 10

9514 1404 393

Answer:

  x = 10·cos(θ) -4·cot(θ)

Step-by-step explanation:

Apparently, we are to assume that the horizontal lines are parallel to each other.

The relevant trig relations are ...

  Sin = Opposite/Hypotenuse

  Cos = Adjacent/Hypotenuse

If the junction point in the middle of AB is labeled X, then we have ...

  sin(θ) = 4/BX   ⇒   BX = 4/sin(θ)

  cos(θ) = x/XA   ⇒   XA = x/cos(θ)

Then ...

  BX +XA = AB = 10

Substituting for BX and XA using the above relations, we get

  4/sin(θ) +x/cos(θ) = 10

Solving for x gives ...

  x = (10 -4/sin(θ))·cos(θ)

  x = 10·cos(θ) -4·cot(θ) . . . . . simplify

_____

We used the identity ...

   cot(θ) = cos(θ)/sin(θ)

9x +14

aisdlkjfhajlsdfhlakjsdhfla

Answer: choice D

Step-by-step explanation:

In order for something to be a function for every x input there must be exactly 1 y output. This means that if x is a number then y must always be 1 number and that 1 number only.

Answer:

The answer is A 1 4/5

Work:

7 1/5- 6 2/5

turn 7 1/5 into 6 6/5 and now subtract the two fractions

6/5-2/5= 4/5

now subtract the whole numbers:

6-6=1

So, your answer should be 4/5 (A)

Answer:

78/2=39°

Step-by-step explanation:

thx for the points

Answer:

Amount paid by insurance company = $6,200

Step-by-step explanation:

Given:

Total cost of car damage = $7,200

Amount deductible = $1,000

Find:

Amount paid by insurance company for total damage

Computation:

Amount paid by insurance company = Total cost of car damage - Amount deductible

Amount paid by insurance company = $7,200 - $1,000

Amount paid by insurance company for total damage = $6,200

Answer: 4) 8

Step-by-step explanation:

-80/10=-8

-8 = - 8. - No

80/-10= - 8. - 8=-8 - No

-(80/10)=-(8)=-8. - 8=-8 - No

8 - Yes.

Answer:

142.7 m²

Step-by-step explanation:

but sure, what you know about triangles and trigonometry (simplified, the relationship of angles and sides of shapes based on circles enclosing them)

so, here a quick summary of important facts and angle/side relationships in triangles :

the sum of all angles in a triangle is always and for every triangle 180 degrees.

sin(angle) is vertically up or down from the horizontal diameter of the encircling circle to the point on the circle hit by a line from the center of the circle with the described angle from the horizontal diameter.

cos(angle) is horizontally left or right from the center of the encircling circle to the point, where the sine-line hits the horizontal diameter.

the sides are named in relation to their opposing corners and their angles. so, for example, side a is opposing the corner point A and the angle at this point (also called A).

here it is also important to know :

a/sin(A) = b/sin(B) = c/sin(C)

and the area of a general triangle is

area = 1/2 × b × c × sin(A) or

1/2 × a × c × sin(B) or

1/2 × a × b × sin(C)

so, if we want to use the first option, we need to calculate b first. for this we use

b/sin(B) = c/sin(C)

b = (c × sin(B)) / sin(C)

for this we need to calculate C first. and we use

A + B + C = 180

C = 180 - A - B = 180 - 54.3 - 30.4 = 95.3 degrees

=>

b = (26.3 × sin(30.4)) / sin(95.3) = 13.36583...

=> area = 1/2 × 13.36583... × 26.3 × sin(54.3) = 142.7324...

= 142.7 (rounded)

Answer:

(121.576 ; 273.424)

Step-by-step explanation:

Given the data:

175, 125, 180, 220, 240, 245

We can calculate the mean and standard deviation

Mean = Σx/ n = 1185 / 6 = 197.5

Standard deviation = 46.125 (calculator)

The confidence interval :

Mean ± margin of error

Margin of Error = Tcritical * s/sqrt(n)

Tcritical at 99%, df = n - 1 ; 6 - 1 = 5

Tcritical = 4.032

Margin of Error = 4.032 * 46.125/√6

Margin of error = 75.924

Confidence interval :

197.5 ± 75.924

Lower boundary = 197.5 - 75.924 = 121.576

Upper boundary = 197.5 + 75.924 = 273.424

(121.576 ; 273.424)

Answer:

this is a linear equation. whatever value you set for x, you substitute and make a table that would probably look like this.

      x      |     y

-------------|-----------

5            | 17/3       <------- what would y be if x is 5?

3             | 5                     lets substitute x with 5

               |                          y=1/3*5+4

               |                         y=5/3+4

               |                        y=17/3

Answer:

0.000064 = 0.0064% probability that the box will contain less than the advertised weight of 466 g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

N(489,6)

This means that [tex]\mu = 489, \sigma = 6[/tex]

What is the probability that the box will contain less than the advertised weight of 466 g?

This is the p-value of Z when X = 466. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{466 - 489}{6}[/tex]

[tex]Z = -3.83[/tex]

[tex]Z = -3.83[/tex] has a p-value of 0.000064

0.000064 = 0.0064% probability that the box will contain less than the advertised weight of 466 g.

Answer:

The appropriate null hypothesis is [tex]H_0: \mu = 22.6[/tex]

The appropriate alternative hypothesis is [tex]H_1: \mu < 22.6[/tex]

Step-by-step explanation:

The average car had a fuel economy of 22.6 MPG. Test if the current average is less than this.

At the null hypothesis, we test if the current average is still of 22.6 MPG, that is:

[tex]H_0: \mu = 22.6[/tex]

At the alternative hypothesis, we test if the current mean has decreased, that is, if it is less than 22.6 MPG. So

[tex]H_1: \mu < 22.6[/tex]

Answer:

Step-by-step explanation:

From the given information:

ΔG° = -30.5 kJ/mol

By applying the following equation to calculate the value of K.

ΔG° =-RT㏑K

making ㏑ K  the subject of the formula:

[tex]\mathtt{ In \ K} = \dfrac{\Delta G^0}{-RT}[/tex]

where;

Temperature at 25° C = (25 + 273)K

= 298K

R = 8.3145 J/mol.K (gas cosntant)

[tex]\mathtt{ In \ K} = \dfrac{-30.5 \times 10^{3}\ J /mol} {-(8.3145 \ J/mol. K \times 298 \ K}[/tex]

[tex]\mathtt{ In \ K} = \dfrac{-30.5 \times 10^{3}\ J /mol} {-2477.721 J/mol }[/tex]

㏑K = 12.309

[tex]K = e^{12.309}[/tex]

K = 221682.17

K = 2.22 × 10⁵

b) The reaction for the metabolism of glucose is given as:

[tex]C_6H_{12} O_6 + 6O_{2(g)} \to + 6CO_{2(g)} + 6H_2O_{(l)}[/tex]

From the above expression, let calculate the Gibbs free energy by using the formula:

[tex]\Delta G^0_{rx n }= \Delta G^0_{product}- \Delta G^0_{reactant}[/tex]

[tex]\Delta G^0_{rx n }= [6 \times \Delta G^0_{f}(CO_2) + 6 \times \Delta G^0_{f}(H_2O)] - [1 \times \Delta G^0_{f}(C_6H_{12}O_6) + 6 \times \Delta G^0_{f}(O_2)][/tex]

At standard conditions;

The values of corresponding compounds are substituted into the equation above:

Thus,

[tex]\Delta G^0_{rx n }= [6 \times (-394) + 6 \times (-237)] - [1 \times (-911) + 6 \times (0)] \ kJ/mol[/tex]

[tex]\Delta G^0_{rx n }= [-2364-1422] - [-911+0] \ kJ/mol[/tex]

[tex]\Delta G^0_{rx n }= -3786 +911 \ kJ/mol[/tex]

[tex]\Delta G^0_{rx n }= -2875 \ kJ/mol[/tex]

[tex]\Delta G^0_{rx n }= -2875000 \ J/mol[/tex]

Now, the no of ATP molecules generated = [tex]\dfrac{\Delta G^0 \text{of metabolism for glucose}}{\Delta G^0 \text{of hydrolysis for ATP}}[/tex]

= (-2875000 J/mol ) / -30500 J/mol

= 94.26

≅ 94 ATP molecules generated

Answer:

600 cm³

Step-by-step explanation:

6000 mm/10 mm = 600 cm

Answer:

B

Step-by-step explanation:

In a boxplot, when the median is in the middle of the box, or when there are equal proportions around the median, the distribution is symmetric. As is shown here, the median is closer to the left side of the box, making it positively skewed (the mean is greater than the median) and not symmetric.

Answer:

m4+m5=180

answer is second option

Answer:

The number of men needed to perform a piece of work twice as great in tenth part of the time  working 8 hours a day supposing that three of the second set can do as much  work as four of the first set is:

Step-by-step explanation:

To find the answer, first, we're gonna find how many hours take to make the piece of work in 16 days, taking into account each day just has 10 hours:

Now, we divide the total hours among the number of persons:

This equivalence isn't the real work of each person, we only need this value to make the next calculations. Now, we have a piece of work twice as great as the first, then, we can calculate the hours the piece of work needs to perform it (twice!):

We need to make this work in tenth part of the time working 8 hours a day, it means:

Now, we know three of the second set can do as much  work as four of the first set, taking into account the calculated equivalence, we have:

So, three persons of the second set can make a equivalence of 8 hours. At last, we calculate all the number of workers we need in a regular time:

Remember we need to perform the job not in a regular time, we need to perform it in tenth part of the time, by this reason, we need 10 times the number of people:

With this calculations, you can find the number of men needed to perform a piece of work twice as great in tenth part of the time  working 8 hours a day supposing that three of the second set can do as much  work as four of the first set is 1200 persons.

Answer:

32

Step-by-step explanation:

14+6(9-6)

[9-6=3]

14+6x3

[6x3=18]

14+18

32

---

hope it helps

Answer:

14+6×(9-6)=32

Answer:

20 degrees

Step-by-step explanation:

Angles on a line equal 180 degrees.

180–140=40

40=2x

x=20

20 degrees

Option B and D are the correct answers

Answer:

A and B I guess so

Answer:

160

Step-by-step explanation:

96 / (100%-40%) = 96/ (60%)

= 96/0.6 = 160

Q) 96/(100% - 40%)

→ 96/ 60%

→ 96/ 0.6

→ 160 is the number.

Answer:

x = (2 + √3) , (2 - √3)

Step-by-step explanation:

GIVEN :-

TO FIND :-

GENERAL FORMULAE TO BE USED IN THIS QUESTION :-

Quadratic formulae -

For a polynomial ax² + bx + c , its roots are :-

[tex]x = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \; ; \frac{-b - \sqrt{b^2 - 4ac} }{2a}[/tex]

SOLUTION :-

Use the quadratic formulae to find the roots of the polynomial.

[tex]=> x = \frac{-(-4) + \sqrt{(-4)^2 - 4 \times 1 \times 1c} }{2 \times 1} \; ; \frac{-(-4) - \sqrt{(-4)^2 - 4 \times 1 \times 1} }{2 \times 1}[/tex]

         [tex]= \frac{4 + \sqrt{16 - 4} }{2} \; ; \frac{4- \sqrt{16 - 4}}{2}[/tex]

         [tex]= \frac{4 + \sqrt{12} }{2} \; ; \frac{4- \sqrt{12}}{2}[/tex]

         [tex]= \frac{4 + 2\sqrt{3} }{2} \; ; \frac{4- 2\sqrt{3}}{2}[/tex]

         [tex]= \frac{2(2 + \sqrt{3})}{2} \; ; \frac{2(2- \sqrt{3})}{2}[/tex]

         [tex]= (2 + \sqrt{3} ) \; ; (2 - \sqrt{3} )[/tex]