Pre Calc work, I need help with writing piece wise functions from graphs. Can anyone help? Giving brainliest

Answer: 46.48 m²

Step-by-step explanation:

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Answer:

5.907 km

Step-by-step explanation:

2.783 km + 3.124 km = 5.907 km

Answer: [tex]160^{\circ}[/tex]

Step-by-step explanation:

By the exterior angle theorem,

[tex]4x-14+5x-182=5x-100\\\\9x-196=5x-100\\\\4x=96\\\\x=24\\\\\implies m\angle BCD=20^{\circ}\\\\\implies m\angle BCA=160^{\circ}[/tex]

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

[tex]\textsf{ \underline{\underline{Steps to solve the problem} }:}[/tex]

By Exterior angle property :

[tex]\qquad❖ \: \sf \:4x - 114 + 5x - 182 = 5x - 100[/tex]

[tex]\qquad❖ \: \sf \:9x -29 6 = 5x - 100[/tex]

[tex]\qquad❖ \: \sf \:9x - 5x = - 100 + 296[/tex]

[tex]\qquad❖ \: \sf \:4x = 96[/tex]

[tex]\qquad❖ \: \sf \:x = 24 \degree[/tex]

Next, Angle BCA = 180° - Angle BCD

( linear pair )

[ let angle BCA = y ]

[tex]\qquad❖ \: \sf \:y = 180 - (5x- 100)[/tex]

[tex]\qquad❖ \: \sf \:y = 180 - (5(24) - 100)[/tex]

[tex]\qquad❖ \: \sf \:y = 180 - (120 - 100)[/tex]

[tex]\qquad❖ \: \sf \:y = 180 - 20[/tex]

[tex]\qquad❖ \: \sf \:y = 160 \degree[/tex]

[tex] \qquad \large \sf {Conclusion} : [/tex]

Answer:

I think the answer is 3.

Step-by-step explanation:

First make arrow diagram of function f and g.

Then show it as composite function of fog(x) in arrow diagram.

Then in range you find -8 whose domain as x is 3.

Answer:

1000

Step-by-step explanation:

The correct answer is Yes. The graph represents a function. The function given is: F(x) = y(-5). Note that the value of x is zero.

A function's graph is the set of all points in the plane of the form (x, f(x)). The graph of f might similarly be defined as the graph of the equation y = f. (x).

Thus, the correct answer or function represented in the graph is f(x) = y(-5)

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well, when it comes to fractions or rationals, they can never have a denominator that's 0, because if that ever happens, the fraction becomes undefined, so the values of "x" or namely the domain values, that we cannot have because they make the fraction undefined are those values that make the denominator 0, we can simply get them by setting the denominator to 0 and check what's "x".

[tex]x^2-9=0\implies x^2=9\implies x=\pm\sqrt{9}\implies x=\pm 3 \\\\[-0.35em] ~\dotfill\\\\ g(x)=\cfrac{x+6}{x^2-9}\hspace{5em} x\ne \begin{cases} 3\\ -3 \end{cases}[/tex]

The number of 150-pound gas cylinders are used each year  for the well is 22.

V = 500 gal/min x 1 year x 525600 min/year

V = 262,800,000 gallons =  994,806,216.835 liters

mass = density x volume

mass = 1.5 mg/L x 994,806,216.835 L

mass = 1,492,209,325 mg = 3289.75 lb

n = 3289.75 lb/150 lb

n = 21.9

n ≈ 22 gas cylinders

Thus, the number of 150-pound gas cylinders are used each year  for the well is 22.

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The coordinate of each point is just the number it corresponds to.

a) 6

b) -5

c) 9

d) 0

The equation of the perpendicular line is [tex]y=-\frac{2}{3}x+4[/tex]

The equation is defined by 2y = 3x + 2

Rewrite the equation in the form y = mx + c

[tex]y=\frac{3}{2}x + 1[/tex]

The slope, m = 3/2

The y-intercept, c = 1

The equation perpendicular to the given line will be of the form:

[tex]y-y_1=\frac{-1}{m}(x-x_1 )[/tex]

Substitute [tex]x_1=3, y_1=2, and m=\frac{3}{2}[/tex] into the equation above

[tex]y-2=\frac{-2}{3} (x-3)\\\\y-2=-\frac{2}{3}x+2\\\\y=-\frac{2}{3}x+4[/tex]

Therefore, the equation of the perpendicular line is [tex]y=-\frac{2}{3}x+4[/tex]

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As you've pointed out, R is a different region that I originally thought. The area of R can be computed using either

[tex]\displaystyle \int_0^2 x^2 \, dx + \int_2^{10} -\frac x2 + 5 \, dx[/tex]

or

[tex]\displaystyle \int_0^4 (10-2y)-\sqrt y \, dy[/tex]

Either way, you've gotten the correct area for part (a).

For part (b), we can use part of the integral with respect to [tex]y[/tex] above. The horizontal distance between the curves [tex]y=x^2[/tex] and [tex]y=-\frac x2+5[/tex] is obtained by first solving for [tex]x[/tex],

[tex]y=x^2 \implies x=\sqrt y[/tex]

[tex]y=-\dfrac x2+5 \implies x = 10-2y[/tex]

so the length of each cross section is [tex]10-2y-\sqrt y[/tex].

The height of each cross section is [tex]3y[/tex].

Then the volume of the solid is

[tex]\displaystyle \int_0^4 3y \left(10-2y-\sqrt y\right) \, dy = \boxed{\int_0^4 -6y^2 + 30y - 3y^{3/2} \, dy}[/tex]

The instructions don't say to evaluate, but if you're looking for practice the volume ends up being 368/5, or 73  3/5.

================================================

Explanation:

These are the steps to focus on

step 3: -6x - 8 < -2

step 4: -6x < 6

The move from the third step to the fourth step has us adding 8 to both sides. Therefore, we use the addition property of inequality.

That property has four forms

It's similar to the idea of starting with a = b, then adding c to both sides to get a+c = b+c

We add the same thing to both sides to keep things balanced.

The number of new cases be on Thursday after all of this growth and falling is 956.

An equation is an expression that shows the relationship between two or more variables and numbers.

There are initially 800 cases, hence:

Number of cases on Monday = 1.15 * 800 = 920

Number of cases on Tuesday = 1.05 * 920 = 966

Number of cases on Wednesday = 0.9 * 966 = 869.4

Number of cases on Thursday = 1.1 * 869.4 = 956.34

The number of new cases be on Thursday after all of this growth and falling is 956.

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The percentage discount from the base price result is 2.16%.

Discount percent will be:

= (Base price - New price)/Base price × 100

= (9.40 - 9.20)/9.40 × 100

= 0.20/9.40 × 100

= 2.16%

The new price will be:

= Base price - (2.16% of base price)

= 9.40 - (2.16% × 9.40)

= 9.20

Therefore, the percentage discount from the base price result is 2.16%.

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Complete question:

A rodent infestation has left a landlord with a major cleanup job; in fact, the cleanup will consist of ripping out all the walls and ceilings of an 1860-square-foot house to replace them. A trip to the local home improvement store reveals that 4'x8' sheets of drywall have three price points as indicated in the table. Another consideration for the landlord is that he is equipped with only a humble half-ton pickup truck, that can hold at most 20 sheets. His truck sips fuel, so he considers only wear and tear on his vehicle at $5 to haul all the sheets he will use for the job. The holding percentage is 20%. The landlord believes the entire job - all walls and ceiling - will require 10,080 square feet of drywall.

Consider the $9.40 per sheet figure as the base price and use the information in Scenario 11.4 to answer the following question. What percentage discount from the base price results in an optimal order quantity of 101 sheets?

1.42%

2.16%

2.03%

1.86%

The probability that it will have an annual snowfall greater than 170 inches is 0.4.

The information is incomplete and the complete question wasn't found. An overview will be given. It should be noted that a probability means the likelihood of the occurence of an event.

Probability is calculated as:

= Number of ways of achieving success/Total number of possible outcomes

Let's assume that the total number of resort is 10 and those that have more than 170 inches are 4.

In this case, the probability will be:

= 4/10

= 0.4

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The area of the triangle which is formed as described is; 60.

According to the task content, when Point A(5,6) is reflected over the y-axis, the point B formed is; (-5,6) and when reflected over both axis, the point C formed is; (-5,-6).

It therefore follows that the length of segments AB and BC which represents the base and height of the triangle formed are; 10 and 12 respectively.

Therefore, the area of the triangle is; (1/2) × 10 × 12 = 60.

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Let [tex]S(t)[/tex] denote the amount of salt (in lbs) in the tank at time [tex]t[/tex] min up to the 10th minute. The tank starts with 100 gal of fresh water, so [tex]S(0)=0[/tex].

Salt flows into the tank at a rate of

[tex]\left(0.5\dfrac{\rm lb}{\rm gal}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = 1\dfrac{\rm lb}{\rm min}[/tex]

and flows out with rate

[tex]\left(\dfrac{S(t)\,\rm lb}{100\,\mathrm{gal} + \left(2\frac{\rm gal}{\rm min} - 2\frac{\rm gal}{\rm min}\right)t}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = \dfrac{S(t)}{50} \dfrac{\rm lb}{\rm min}[/tex]

Then the net rate of change in the salt content of the mixture is governed by the linear differential equation

[tex]\dfrac{dS}{dt} = 1 - \dfrac S{50}[/tex]

Solving with an integrating factor, we have

[tex]\dfrac{dS}{dt} + \dfrac S{50} = 1[/tex]

[tex]\dfrac{dS}{dt} e^{t/50}+ \dfrac1{50}Se^{t/50} = e^{t/50}[/tex]

[tex]\dfrac{d}{dt} \left(S e^{t/50}\right) = e^{t/50}[/tex]

By the fundamental theorem of calculus, integrating both sides yields

[tex]\displaystyle S e^{t/50} = Se^{t/50}\bigg|_{t=0} + \int_0^t e^{u/50}\, du[/tex]

[tex]S e^{t/50} = S(0) + 50(e^{t/50} - 1)[/tex]

[tex]S = 50 - 50e^{-t/50}[/tex]

After 10 min, the tank contains

[tex]S(10) = 50 - 50e^{-10/50} = 50 \dfrac{e^{1/5}-1}{e^{1/5}} \approx 9.063 \,\rm lb[/tex]

of salt.

Now let [tex]\hat S(t)[/tex] denote the amount of salt in the tank at time [tex]t[/tex] min after the first 10 minutes have elapsed, with initial value [tex]\hat S(0)=S(10)[/tex].

Fresh water is poured into the tank, so there is no salt inflow. The salt that remains in the tank flows out at a rate of

[tex]\left(\dfrac{\hat S(t)\,\rm lb}{100\,\mathrm{gal}+\left(2\frac{\rm gal}{\rm min}-2\frac{\rm gal}{\rm min}\right)t}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = \dfrac{\hat S(t)}{50} \dfrac{\rm lb}{\rm min}[/tex]

so that [tex]\hat S[/tex] is given by the differential equation

[tex]\dfrac{d\hat S}{dt} = -\dfrac{\hat S}{50}[/tex]

We solve this equation in exactly the same way.

[tex]\dfrac{d\hat S}{dt} + \dfrac{\hat S}{50} = 0[/tex]

[tex]\dfrac{d\hat S}{dt} e^{t/50} + \dfrac1{50}\hat S e^{t/50} = 0[/tex]

[tex]\dfrac{d}{dt} \left(\hat S e^{t/50}\right) = 0[/tex]

[tex]\hat S e^{t/50} = \hat S(0)[/tex]

[tex]\hat S = 50 \dfrac{e^{1/5}-1}{e^{1/5}} e^{-t/50}[/tex]

After another 10 min, the tank has

[tex]\hat S(10) = 50 \dfrac{e^{1/5}-1}{e^{1/5}} e^{-1/5} = 50 \dfrac{e^{1/5}-1}{e^{2/5}} \approx \boxed{7.421}[/tex]

lb of salt.

There are 7 pounds in 112 ounces

There is 1 pound in 16 ounces

So, the number of pounds in 112 ounces is

Pounds = 116 ounces/16 ounces

Remove the unit

Pounds = 112/16

Evaluate the quotient

Pounds = 7

Hence, there are 7 pounds in 112 ounces

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Using the Factor Theorem, the function factored into linear factors is given as follows:

[tex]3x^2 + 23x^2 - 35x + 9 = 3(x + 9)(x - 1)\left(x - \frac{1}{3}\right)[/tex]

The Factor Theorem states that a polynomial function with roots [tex]x_1, x_2, \codts, x_n[/tex] is given by:

[tex]f(x) = a(x - x_1)(x - x_2) \cdots (x - x_n)[/tex]

In which a is the leading coefficient.

-9 is a zero of f(x), hence [tex]x_1 = -9[/tex] and:

3x³ + 23x² - 35x + 9 = (x + 9)(ax² + bx + c)

ax³ + (b + 9a)x² + (9b + c)x + 9c = 3x³ + 23x² - 35x + 9

The coefficients are given as follows:

Hence:

3x³ + 23x² - 35x + 9 = (x + 9)(3x² - 4x + 1)

[tex]3x^2 + 23x^2 - 35x + 9 = 3(x + 9)(x - 1)\left(x - \frac{1}{3}\right)[/tex]

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The probability is P = 0.135

First, we need to find the individual probabilities, that are given by the quotient between the number of outcomes that meet the condition and the total number of outcomes.

The joint probability is given by the product between the individual probabilities, so we get:

P = (1/2)*(3/6)*(28/52) = 0.135

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Using translation concepts, it is found that the point (0, 0) on the graph of f(x) corresponds to point (4,-7) on g(x).

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

Considering the translations, the equivalent point is found as follows:

Hence the point (0, 0) on the graph of f(x) corresponds to point (4,-7) on g(x).

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The expression for the total area of the wall is 10x.

What is area of rectangle?

Let l be length and b be breadth of the rectangle.

Then the area is the product of length and breadth of the rectangle.

That is area = l* b

We can find the expression for the total area of the wall as shown below:

Let the total length of the wall be x.

Height=10 ft

Total area = Height*Length

Putting the value of length and breadth in the above formula.

Total area = 10*x

Total area = 10x

Hence, the expression for the total area of the wall is 10x.

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The point that is on the graph of the function [tex]F(x) = 5^x[/tex] is given by:

D. (1,5).

The function is defined by:

[tex]F(x) = 5^x[/tex]

When x = 0, [tex]F(x) = 5^0 = 1[/tex], hence point (0,1) is on the graph of the function.

When x = 1, [tex]F(x) = 5^1 = 5[/tex], hence point (1,5) is on the graph of the function, which means that option D is correct.

When x = 5, [tex]F(x) = 5^5 = 3125[/tex], hence point (5,3125) is on the graph of the function.

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The complete proof for part D is given in the attached text. Hence, by the nature of the converse corresponding angle,

OP is parallel to MN. (OP ║ MN).

A mathematical proof is an argument that is inferential with respect to a mathematical assertion that demonstrates that the provided assumptions logically ensure the conclusion.

The complete proof for part D is given as follow:

If a transversal line "t" divides through OP and MN as given in the attached image, and SE and "T.F" are the divider of ∠QSD and ∠STM, then:

∠QSE = 1/2 ∠QSO; and

∠"STF" = 1/2 ∠STM

If the corresponding angle is ∠QST = "STF", then

QSD = "STF"

Hence, by the nature of the converse corresponding angle,

OP is parallel to MN

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Answer:

y=5

Step-by-step explanation:

OK, let see, 2x+y=7, –3x+y=2:

[tex]2x + y = 7 \\ - 3x + y = 2 \\ we \: need \: to \: solve \: the \: equation \: so \\ - 2x - y = - 7 \\ - 3x + y = 2 \\ - 5x = - 5 = = > x = 1 \\ 2(1) + y = 7 = = = > y = 5[/tex]

The solution to the Questions are

(a)

The alternate hypothesis demonstrates that there are two possible outcomes for the test.

(b)

Here we have

[tex]n_{1}=40,\\\\ \bar{x}_{1}=102,\sigma_{1}=5,n_{2}=50,\bar{x}_{2}=99,\sigma_{2}=6[/tex]

(b)

Here the test is two-tailed. So for [tex]\alpha =0.04[/tex], the critical values of the z-test are -2.05 and 2.05.

Decision rule: If z > 2.05 or z<-2.05, reject H0

(c)

Test statistics will be

[tex]z=\frac{(\bar{x}_{1}-\bar{x}_{2})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}} \\\\\\z=\frac{(102-99)-(0)}{\sqrt{\frac{5^{2}}{40}+\frac{6^{2}}{50}}}[/tex]

z=2.59

(d)

The two-tailed nature of the test is shown by the alternative hypothesis.

(e)

The result of the test yields the following P-value: 0.0096

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Based on the amounts that Janet Lopez has to invest in stocks and bonds, the linear programming model would be:

The return on stocks (x) is 18% and the return on bonds (y) is 6%, The objective function:

= 0.18x + 0.06y

There are constraints to watch out for:

Maximum to lose on stocks is 22% and on bonds is 5% but these are to be less than the total amount of $100,000.

0.22x + 0.05y ≤ 100,000

The total amount to invest is $72,000 which means that both bonds and stocks need to be less than this amount:

x + y ≤ 720,000

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The probability that a student attended the game, given that the student is from North Beach is 0.65.

Probability determines the chances that an event would happen. The probability the event occurs is 1 and the probability that the event does not occur is 0.

The probability that a student attended the game, given that the student is from North Beach = number of North beach students that attended the game / total number of students who attended the game

110 / 170 = 0.65

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