Does this graph represent a function? Why or why not?

Inequalities help us to compare two unequal expressions. The value of x must be less than -19/13.

Inequalities help us to compare two unequal expressions. Also, it helps us to compare the non-equal expressions so that an equation can be formed. It is mostly denoted by the symbol <, >, ≤, and ≥.

The given inequalities can be simplified as,

1.)

−9x+2>18

−9x>18-2

−9x>16

x < -16/9

Thus, the value of x must be less than -16/9.

2.)

13x+15≤−4

13x ≤ -19

x ≤ -19/13

Thus, the value of x must be less than -19/13.

For the given condition −9x+2>18 OR 13x+15≤−4, the value of x must follow any one of the inequality. Hence, the value of x must be less than -19/13.

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Answer:

1A) 2 gallons of 20% solution and 3 gallons 15% solution needed

1B) 4 gallons of 20% solution and 1 gallons 15% solution needed

Step-by-step explanation:

1A) adding 20% salt and 15% water making 5 gallons of 17%

20% salt + 15% salt = 5 gallons of 17% salt

0.20x + 0.15(5 - x) = 0.17(5)

0.20x + 0.75 - 0.15x = 0.85

0.20x + 0.75 - 0.75 - 0.15x = 0.85 - 0.75

0.20x - 0.15x = 0.10

0.05x = 0.10

0.05x/ 0.05 = 0.10/0.05

x = 2 gallons (amount of 20% solution needed)

5 - x = 5 - 2 = 3 gallons (amount of 15% solution needed)

1B)

0.20x + 0.15(5 - x) = 0.19(5)

0.20x + 0.75 - 0.15x = 0.95

0.20x + 0.75 - 0.75 - 0.15x = 0.95 - 0.75

0.20x - 0.15x = 0.20

0.05x = 0.20

0.05x / 0.05 = 0.20/0.05

x = 4 gallons (amount of 20% solution needed)

5 - x = 5 - 4 = 1 gallons (amount of 15% solution needed)

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The expression of 2B - 3C in standard form is -3n^2 + 24n - 2

Given the following expressions

B = 3n - 10

C = n^2-6n-6

Required

2B - 3C

Substitute

2B - 3C = 2(3n -10) - 3(n^2-6n-6)

2B - 3C = 6n - 20 -3n^2+18n+18

2B - 3C = -3n^2 + 24n - 2

Hence expression of 2B - 3C in standard form is -3n^2 + 24n - 2

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Answer: (99/13, 150/13)

Step-by-step explanation:

I'm too lazy to explain so here's a screen shot from desmos.

The statistical test which can be used to analyze the data is chi square test.

Given two categories of tomatoes , one is red and other one is green.

When we want to compare more than two categorical variables, we can use the chi-square test. The main objective of chi square test is to compare the distribution of responses or the proportions of participants in each response category. This test does not require mean, standard deviation or anything. In the given problem we have not given sample mean, population mean, population standard deviation, etc. So we cannot use z test, t test, f test because they require sample mean, population mean , standard deviation in their calculation. The formula to be use in chi square test is as under:

[tex]X^{2}=[/tex]∑[tex](f_{0} -f_{e}) ^{2} /f_{e}[/tex]

where [tex]f_{0}[/tex]= observed frequency of each of response category.

[tex]f_{1}[/tex]=expected frequency in each of the response categories.

We can find the critical value in a table of probabilities for the chi square distribution with degree of freedom df=K-1.

Hence we can use chi square test to analyze the data.

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Answer:

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

Add 2 to both sides of equation:

Then follow the steps:

Take square root of both sides to get:

We are taking the positive value as we are told x > 0, hence the answer is 7.

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

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

[tex] \qquad❖ \: \sf \: {x}^{2} + \cfrac{1}{ {x}^{2} } = 47[/tex]

[tex] \qquad❖ \: \sf \: {x}^{2} + \cfrac{1}{ {x}^{2} } + 2 - 2 = 47[/tex]

( adding and subtracting 2, doesn't change the value )

[tex] \qquad❖ \: \sf \: {x}^{2} + \cfrac{1}{ {x}^{2} } + 2 = 47 + 2[/tex]

[tex] \qquad❖ \: \sf \: {x}^{2} + \cfrac{1}{ {x}^{2} } + \bigg(2 \sdot x \sdot \cfrac{1}{x} \bigg) = 49[/tex]

( x cancel out, so no change in value )

[tex] \qquad❖ \: \sf \: \bigg(x + \cfrac{1}{x} \bigg) {}^{2} = 49[/tex]

( use identity a² + 2ab + b² = (a + b)² )

[tex] \qquad❖ \: \sf \: \bigg(x + \cfrac{1}{x} \bigg) {}^{} = \sqrt{49} [/tex]

[tex] \qquad❖ \: \sf \: x + \cfrac{1}{x} {}^{} = \pm7[/tex]

since we have to take positive value, i.e greater than 0

[tex] \qquad❖ \: \sf \: x + \cfrac{1}{x} {}^{} = 7[/tex]

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

Therefore, the required value is 7

Answer:

Step-by-step explanation:

Answer:

True

Step-by-step explanation:

If ab < 0, then ab = negative #.

In order for ab to be a negative #, one of them has to be negative while the other one needs to be positive.

Example:

a = -2, b = 1

ab < 0

(-2)(1) < 0

-2 < 0, TRUE

a < 0

-2 < 0, TRUE

b > 0

1 > 0, TRUE

If I switch a = -2 to 1 and b = 1 to -2, a > 0 and b < 0 is true too.

The description below proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.

Let ABC be an isosceles triangle such that AB = AC.

Let AD be the bisector of ∠A.

We want to prove that BD=DC

In △ABD & △ACD

AB = AC(Thus, △ABC is an isosceles triangle)

∠BAD =∠CAD(Because AD is the bisector of ∠A)

AD = AD(Common sides)

By SAS Congruency, we have;

△ABD ≅ △ACD

By corresponding parts of congruent triangles, we can say that; BD=DC

Thus, this proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.

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Answer:

The line has a downward slope

Step-by-step explanation:

If y = ax, it means that a represents the slope of the line. So, if A is negative, the line has a downward slope. In other words, for every increase in the value of x, there is a decrease in the value of y (increase in the negative direction). You can imagine the line y = -x, where the value of a is -1. When x is 1, y is -1; when x is 2, y is -2; when x is 5, y is -5. You can imagine that the graph looks like a diagonal line, much like y = x, except the function values are positive in the second quadrant and negative in the fourth (it slants downward to the right side). I hope that answers your question.

Answer:

9x^3+6x^2+21x-8

Step-by-step explanation:

(3x-1) (3x ^ 2 + 3x + 8) =3x(3x ^ 2 + 3x + 8) +(-1)(3x ^ 2 + 3x + 8)  =9x^3+9x^2+24x-3x^2-3x-8 = 9x^3+6x^2+21x-8

Answer:

[tex]\huge\boxed{\sf 9x^3+6x^2+21x-8}[/tex]

Step-by-step explanation:

= [tex](3x-1)(3x^2+3x+8)[/tex]

[tex]Expand[/tex]

[tex]= 3x(3x^2+3x+8)-1(3x^2+3x+8)\\\\Multiply\\\\=9x^3+9x^2+24x-3x^2-3x-8\\\\Combine \ like \ terms\\\\= 9x^3+9x^2-3x^2+24x-3x-8\\\\= 9x^3+6x^2+21x-8\\\\\rule[225]{225}{2}[/tex]

Answer:

Step-by-step explanation:

Answer:10.2km

He took the route eastwards then southwards

Answer:

y-intercept = -3

another point on the line = (6,-1)

Step-by-step explanation:

1. Underline important parts of the question (slope of 1/3, point (3, -2))

2. mark the point that is given (3, -2)

3. slope form is rise over run which basically means that the number one is how much the line rises, and the number three is how much the line goes to the right (since the slope is positive then it runs to the right, if it was negative it would go to the left.) Basically 1/3 = rise over run

4. put your pencil on the point (3, -2) and go up one point and over three points. continue doing so from each new point (two or three points is good enough)

5. take a ruler and draw a line that connects all three and keep it going until it crosses the y- axis (which is the vertical line)

6. the point that crosses through the y -axis is the y-intercept.

7. now you can go and look at your points that you have made and choose one of them to answer the question with

(I hope this cleared it up a bit I'm not very good at explaining things this is the best I can do)

There are

[tex]\dbinom{2n}2 = \dfrac{(2n)!}{2! (2n-2)!}[/tex]

ways of pairing up any 2 members from the pool of [tex]2n[/tex] contestants. Note that

[tex](2n)! = 1\times2\times3\times4\times\cdots\times(2n-2)\times(2n-1)\times(2n) = (2n-2)! \times(2n-1) \times(2n)[/tex]

so that

[tex]\dbinom{2n}2 = \dfrac{(2n)\times(2n-1)\times(2n-2)!}{2! (2n-2)!} = \boxed{n(2n-1)}[/tex]

The number of different ways in which first-round matches can be conducted is n (2n - 1).

We have,

Number of contestant = 2n

Number of contestants in each match = 2

Now,

The number of different ways in which first-round matches can be conducted.

A combination formula is used.

= [tex]^{2n}C_2[/tex]

= 2n! / 2! (2n - 2)!

= 2n (2n - 1)(2n - 2)! / [2 x (2n - 2)!]

= 2n (2n - 1) / 2

= n (2n - 1)

Thus,

The number of different ways in which first-round matches can be conducted is n (2n - 1).

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The expression that represents the width is w = (P-2l)/2

Given the formula for calculating the perimeter of a triangle expressed as:

P = 2(l + w)

where

l is the length

w is the width

Make w the subject of the formula

Given

P = 2(l + w)

Expand

P = 2l + 2w

Subtract 2l from both sides

P - 2l = 2w

Divide both sides by2

2w/2 = (P-2l)/2

w = (P-2l)/2

Hence the expression that represents the width is w = (P-2l)/2

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Answer:

Answer is 32

Step-by-step explanation:

150-22=128

128÷4=32

Answer:

0.5

one divide by two is zero point five

[tex]\frak{Hi!}[/tex]

[tex]\orange\hspace{300pt}\above2[/tex]

                   We need to find the decimal form of [tex]\boldsymbol{\sf{\displaystyle\frac{1}{2}}}[/tex].

                  But first we need to convert it into a fraction whose

                   denominator is 10, or 100, or 1,000, etc.

                   The closest number that's divisible by 2 is 10.

                   So that's the new denominator of the fraction.

                   Now we should also multiply the top times the number

                   that we multiplied the bottom by. See how this works?

                   [tex]\boldsymbol{\sf{\displaystyle\frac{1\times5}{2\times5}}}[/tex]. This yields

                  [tex]\boldsymbol{\sf{\displaystyle\frac{5}{10}}}[/tex]. And 5/10 in decimal-form yields

                  [tex]\boldsymbol{\sf{0.5}}[/tex].

               [tex]\orange\hspace{300pt}\above3[/tex]  

               [tex]\LARGE\boldsymbol{\sf{calligraphy}}[/tex]

Answer:

Step-by-step explanation:

(i) Given that

[tex]\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}[/tex]

when [tex]x=1[/tex] this reduces to

[tex]\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}[/tex]

[tex]\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}[/tex]

[tex]2 \tan^{-1}(y) = \dfrac\pi3[/tex]

[tex]\tan^{-1}(y) = \dfrac\pi6[/tex]

[tex]\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)[/tex]

[tex]\implies \boxed{y = \dfrac1{\sqrt3}}[/tex]

(ii) Differentiate [tex]\tan^{-1}(xy)[/tex] implicitly with respect to [tex]x[/tex]. By the chain and product rules,

[tex]\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}[/tex]

(iii) Differentiating both sides of the given equation leads to

[tex]\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0[/tex]

where we use the result from (ii) for the derivative of [tex]\tan^{-1}(xy)[/tex].

Solve for [tex]\frac{dy}{dx}[/tex] :

[tex]\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0[/tex]

[tex]\left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} = -\left(\dfrac1{1+x^2} + \dfrac y{1+x^2y^2}\right)[/tex]

[tex]\dfrac{1+x^2y^2 + x(1+y^2)}{(1+y^2)(1+x^2y^2)} \dfrac{dy}{dx} = - \dfrac{1+x^2y^2 + y(1+x^2)}{(1+x^2)(1+x^2y^2)}[/tex]

[tex]\implies \dfrac{dy}{dx} = - \dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2) (1 + x^2y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2) (1+x^2y^2)}[/tex]

[tex]\implies \dfrac{dy}{dx} = -\dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2)}[/tex]

From part (i), we have [tex]x=1[/tex] and [tex]y=\frac1{\sqrt3}[/tex], and substituting these leads to

[tex]\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}[/tex]

[tex]\dfrac{dy}{dx} = -\dfrac{\left(\frac43 + \frac2{\sqrt3}\right) \times \frac43}{\frac83 \times 2}[/tex]

[tex]\dfrac{dy}{dx} = -\dfrac13 - \dfrac1{2\sqrt3}[/tex]

as required.

The area of the shaded region shown in the figure is 80π unit²

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

Area of the shaded region = area of semicircle IL - area of semicircle JK + area of semicircle IJ + area of semicircle KL

Area of the shaded region = area of semicircle IL + area of semicircle IJ = (0.5 * π * 12²) + (0.5 * π * (12/3)²) = 80π

The area of the shaded region shown in the figure is 80π unit²

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Answer:

Step-by-step explanation:

You need to find the set of points that will yield a slope that is the negative reciprocal of the slope of Line L because perpendicular lines have negative reciprocal slopes. The negative reciprocal of 13/7 is -7/13. Which set of points will produce this result? The formula for finding the slope is:

m = (y2 - y1)/(x2 - x1)

Consider the second set of coordinates.

(2 - (-5))/(-7 - 6) = (2 + 5)/(-13) = -7/13

The second set of coordinates satisfy the condition.

Answer:

No, it will not pass specifications.

Step-by-step explanation:

It says that the max slope of the wheelchair ramp should be 1/12.

To calculate slope, you do rise over run or y values over x values. In this case, the rise over run is 5/50. 5/50 simplifies to 1/10.

1/10 > 1/12, meaning that it is more than the maximum slope.

Answer:

1/2

Step-by-step explanation:

The dimensions are cut in half ....scale factor 1/2

Answer:

200miles

Step-by-step explanation

4 x 50 = 200

As for the table, just multiply each inch by 50.

For example; 7in = 350 miles because 7x50 = 350

Answer:

Step-by-step explanation:

Comment

You don't have to do anything to the 6 feet. It already is in feet. It's the 2 inches you have to worry about.

2 inches = 2/12 of a foot.

2/12 = 1/6 = 0.167 feet

Answer

6 feet 2 inches = 6.167  feet

Step-by-step explanation:

Answer:

[tex]x=10^{4}[/tex]

Step-by-step explanation:

An exponential base is the inverse of a logarithm with the same base.  Given the equation log x = 4, note that the base of the logarithm isn't written in (it has no subscript).  By default, the base of the logarithm function is base-10.  So, to rewrite the equation using an exponential, we need to undo the logarithm with an exponential of the same base.

[tex]\log(x)=4[/tex]

[tex]10^{\log{(x)}}=10^{4}[/tex]

[tex]x=10^{4}[/tex]

Think about the equation x + 7 = 10.

I could tell you this is the equation in "addition form".  What if I asked you to write an equation in "subtraction form"?  While neither "addition form" nor "subtraction form" have been defined explicitly, one could undo the addition, and get an equation with subtraction in it, which one could argue is a "subtraction form" of the equation.

[tex]x+7=10[/tex]

Subtracting 7 from both sides to undo the addition...

[tex](x+7)-7=(10)-7[/tex]

Simplifying the left side since the "+7" and "-7" cancel...

[tex]x=10-7[/tex]

This is arguably in a "subtraction form" (there's a subtraction in it), whereas the original equation had addition in it.

While we could simplify this particular equation's right-hand side, we might not always be able to (what if the 10 had been a "y"... then the "y" and the "7" aren't like terms, and they would have to remain separate)

Similarly, in the logarithm problem, one could simplify 10^4 (it's 10,000), but one doesn't have to, and one won't always be able to.

For the logarithm problem, x=10^4 is the exponential form, as requested.

Answer:

Option C

Step-by-step explanation:

We can see that these two triangles are similar triangles. And based on the sides we also know triangle ZXY and triangle FDE have a ratio of 3:2. We can find side x by the equation:

3:2=9:x

3x=18

x=6 or option C

Answer:

C.6

Step-by-step explanation:

12x=8

x=8/12=2/3

2/3*9=18/3=6

Answer:

Yes, it is a linear equation.

Step-by-step explanation:

   An equation of the form ax + by +c = 0, where x and y are two variables and a, b,c are the non-zero real numbers is called a linear equation and the degree of the equation is 1.

       y = 1 - x

x + y - 1 = 0

Where a = 1 ; b=1 and c = -1

So, y = 1 -x is a linear equation.

There are two(2) possible solutions for the given parameters of a triangle. Using the law of cosines, the parameters of the triangle are calculated.

The law of cosines is used to relate the angles and side lengths of a triangle.

They are given as follows:

a² = b² + c² -2bc cos(A)

b² = a² + c² - 2ac cos(B)

c² = a² + b² - 2ab cos(C)

The given triangle has B = 37°, a = 32 and b = 27

So, using the law b² = a² + c² - 2ac cos(B)

On substituting,

27² = 32² + c² - 2(32)c cos(37°)

⇒ c² - 50.56 c + 1024 = 729

⇒ c² - 50.56 c + 295 = 0

Since the obtained equation is quadratic, it has 2 solutions.

Using quadratic formula the solutions are calculated.

c = [50.56 ± [tex]\sqrt{50.56^2-4(1)(295)}[/tex]]/2(1)

On simplifying,

c = 43.8 and c = 6.7

Therefore, there are 2 possible solutions for the given triangle.

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