What is the slope of the line through the points (4,2) and (-16,-6)

Answer:

[tex]y\ =\ \left|\frac{1}{2}x-2.5\right|+3[/tex]

Step-by-step explanation:

Look at the image below ↓

Answer:

Check below the graph.

Step-by-step explanation:

Hi, for this function, check the graph below:

1) Note that in this function the value outside the brackets points how high the graph will be traced.

2) The value within the brackets, points since it's a negative expression how far to the right the graph will be traced.

Answer:

83

Step-by-step explanation:

You're given two vertical angles, and vertical angles are congruent. This means that (6x - 7) = (4x + 23); x = 15. Plug it into ABC (which is (6x - 7)) to get 6(15) - 7 = 90 - 7 = 83

Answer:

2x⁴+x³-x²+54x+56

Step-by-step explanation:

Given the expression length of dylan room =  (x² – 2x + 8) and width = (2x² + 5x – 7), assuming the shap of the room is rectangular in nature, the formula for calculating area of a triangle is given as;

Area of rectangle = Length *Width

Area of the rectangle =  (x² – 2x + 8)(2x² + 5x – 7)

Area of the rectangle  = x²(2x² + 5x – 7) - 2x (2x² + 5x – 7) + 8(2x² + 5x – 7)

= (2x⁴+5x³-7x²)-(4x³+10x²-14x)+(16x²+40x-56)

expanding the bracket

= 2x⁴+5x³-7x²-4x³-10x²+14x+16x²+40x-56

Collecting the like terms;

= 2x⁴+5x³-4x³-7x²-10x²+16x²+40x+14x+56

= 2x⁴+x³-x²+54x+56

Hence, the expression that represents the area (lw) of Dylan's room is 2x⁴+x³-x²+54x+56

Answer:

2x^4+ x^3 - x^2 + 54x - 56 expression represents the area of Dylan’s room

Step-by-step explanation:

C on edge :)

Answer: If a scatterplot is included in the assignment

The dots plotted on the graph might closely follow the graph of exponential decline. There is a large number of texts per day by 19-20-21 year-olds, but the number seems to decline exponentially as age increases. With a little work, it may be possible to plot the curve and write an equation to model the decline.

Step-by-step explanation:  Look at some graphs of exponential decay. Also consider harmonic and hyperbolic decay. The trend in the data is evident. The main challenge is to look at the data and create an equation that models it.

By the factor theorem,

[tex]3x^2+5x+7=3(x-u)(x-v)\implies\begin{cases}uv=\frac73\\u+v=-\frac53\end{cases}[/tex]

Now,

[tex](u+v)^2=u^2+2uv+v^2=\left(-\dfrac53\right)^2=\dfrac{25}9[/tex]

[tex]\implies u^2+v^2=\dfrac{25}9-\dfrac{14}3=-\dfrac{17}9[/tex]

So we have

[tex]\dfrac uv+\dfrac vu=\dfrac{u^2+v^2}{uv}=\dfrac{-\frac{17}9}{\frac73}=\boxed{-\dfrac{17}{21}}[/tex]

The value of [tex]\frac{u}{v} +\frac{v}{u}[/tex] is [tex]\frac{-17}{21}[/tex].

A quadratic equation is an algebraic equation of the second degree in x. The quadratic equation in its standard form is[tex]ax^{2} +bx+c=0[/tex], where a and b are the coefficients, x is the variable, and c is the constant term.

If [tex]ax^{2} +bx+c = 0[/tex] be the quadratic equation then

Sum of the roots = [tex]\frac{-b}{a}[/tex]

And,

Product of the roots = [tex]\frac{c}{a}[/tex]

According to the given question.

We have a quadratic equation [tex]3x^{2} +5x+7=0..(i)[/tex]

On comparing the above quadratic equation with standard equation or general equation [tex]ax^{2} +bx+c = 0[/tex].

We get

[tex]a = 3\\b = 5\\and\\c = 7[/tex]

Also, u and v are the solutions of the quadratic equation.

⇒ u and v are the roots of the given quadratic equation.

Since, we know that the sum of the roots of the quadratic equation is [tex]-\frac{b}{a}[/tex].

And product of the roots of the quadratic equation is [tex]\frac{c}{a}[/tex].

Therefore,

[tex]u +v = \frac{-5}{3}[/tex] ...(ii) (sum of the roots)

[tex]uv=\frac{7}{3}[/tex]   ....(iii)       (product of the roots)

Now,

[tex]\frac{u}{v} +\frac{v}{u} = \frac{u^{2} +v^{2} }{uv} = \frac{(u+v)^{2}-2uv }{uv}[/tex]                    ([tex](a+b)^{2} =a^{2} +b^{2} +2ab[/tex])

Therefore,

[tex]\frac{u}{v} +\frac{v}{u} =\frac{(\frac{-5}{3} )^{2}-2(\frac{7}{3} ) }{\frac{7}{3} }[/tex]         (from (i) and (ii))

⇒ [tex]\frac{u}{v} +\frac{v}{u} =\frac{\frac{25}{9}-\frac{14}{3} }{\frac{7}{3} }[/tex]

⇒ [tex]\frac{u}{v} +\frac{v}{u} = \frac{\frac{25-42}{9} }{\frac{7}{3} }[/tex]

⇒ [tex]\frac{u}{v} +\frac{v}{u} = \frac{\frac{-17}{9} }{\frac{7}{3} }[/tex]

⇒ [tex]\frac{u}{v} +\frac{v}{u} =\frac{-17}{21}[/tex]

Therefore, the value of [tex]\frac{u}{v} +\frac{v}{u}[/tex] is [tex]\frac{-17}{21}[/tex].

Find out more information about sum and product of the roots of the quadratic equation here:

https://brainly.com/question/14266582

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Answer:

pretty sure this is right

Answer:

b > 3 2/15

Step-by-step explanation:

To make it easier to solve convert the mixed fraction to a fraction.

2 3/5 = 13/5

Now, multiply the fraction by 3/3 so that you will have a common denominator.

13/5 × 3/3 = 39/15

Now you solve for b.

39/15 < b - 8/15

39/15 + 8/15 < (b - 8/15) + 8/15

47/15 < b

b > 47/15

Convert the fraction to a mixed fraction to find the answer

47/15 = 3 2/15

b > 3 2/15

Answer:

Step-by-step explanation:

In this problem we are going find the natural logarithmic of the numbers involved and solve for x

[tex]ln65-Ln x= 39\\[/tex]

from tables

[tex]4.17-ln x= 39\4.17-39= lnx\\-34.83=lnx\\[/tex]

taking the exponents of both sides we have

[tex]e^-^3^4^.^8^3= x\\x= 7.47[/tex]

Answer:

10a²b²6ab²

Step-by-step explanation:

Distribute the 2ab the other values

Answer:

[tex]\displaystyle \sin \theta = \frac{4}{5}[/tex] if [tex]\displaystyle \cos\theta = -\frac{3}{5}[/tex] and [tex]\theta[/tex] is in the second quadrant.

Step-by-step explanation:

By the Pythagorean Trigonometric Identity:

[tex]\left(\sin \theta\right)^2 + \left(\cos\theta)^2 = 1[/tex] for all real [tex]\theta[/tex] values.

In this question:

[tex]\displaystyle \left(\cos\theta\right)^2 = \left(-\frac{3}{5}\right)^2 = \frac{9}{25}[/tex].

Therefore:

[tex]\begin{aligned} \left(\sin\theta\right)^2 &= 1 -\left(\cos\theta\right)^2 \\ &= 1 - \left(\frac{3}{5}\right)^2 = \frac{16}{25}\end{aligned}[/tex].

Note, that depending on [tex]\theta[/tex], the sign [tex]\sin \theta[/tex] can either be positive or negative. The sine of any angles above the [tex]x[/tex] axis should be positive. That region includes the first quadrant, the positive [tex]y[/tex]-axis, and the second quadrant.

According to this question, the [tex]\theta[/tex] here is in the second quadrant of the cartesian plane, which is indeed above the [tex]x[/tex]-axis. As a result, the sine of this

It was already found (using the Pythagorean Trigonometric Identity) that:

[tex]\displaystyle \left(\sin\theta\right)^2 = \frac{16}{25}[/tex].

Take the positive square root of both sides to find the value of [tex]\sin \theta[/tex]:

[tex]\displaystyle \sin\theta =\sqrt{\frac{16}{25}} = \frac{4}{5}[/tex].

Answer:

Series A has an r value of 2/5 and series A has an r value of 3. The sum of the series A is 50/3

Step-by-step explanation:

A geometric sequence is in the form a, ar, ar², ar³, .  .   .

Where a is the first term and r is the common ratio = [tex]\frac{a_{n+1}}{a_n}[/tex]

For series A:  10+4+8/5+16/25+32/125+⋯   The common ratio r is given as:

[tex]r=\frac{a_{n+1}}{a_n}=\frac{4}{10} =\frac{2}{5}[/tex]

For series B: 1/5+3/5+9/5+27/5+81/5+⋯   The common ratio r is given as:

[tex]r=\frac{a_{n+1}}{a_n}=\frac{3/5}{1/5} =3[/tex]

For series A a = 10, r = 2/5, which mean 0 < r < 1, the sum of the series is given as:

[tex]S_{\infty}=\frac{a}{1-r}=\frac{10}{1-\frac{2}{5} } =\frac{50}{3}[/tex]

Answer:

Yes, the ordered pair is correct.

Explanation:

You can check the if the ordered pair by substituting the values into the equation. If you substitute the ordered pair (1, 3), then you can make sure the ordered pair is correct. The equation with the substitution will be 3 = 1 + 2, which results in the true equation 3 = 3, therefore the ordered pair is correct.

Answer:

  x = 5 is the solution

Step-by-step explanation:

See the attachment for table values.

  f(x) = g(x) for x = 5

The solution to the equation ...

  2.5x -10.5 = 64(0.5^x)

is x = 5.

Answer:

f(x) = 4x² + 48x + 149

Step-by-step explanation:

Given

f(x) = 4(x + 6)² + 5 ← expand (x + 6)² using FOIL

     = 4(x² + 12x + 36) + 5 ← distribute parenthesis by 4

     = 4x² + 48x + 144 + 5 ← collect like terms

     = 4x² + 48x + 149 ← in standard form

Answer:

[tex]f(x)=4x^{2} +149[/tex]

Step-by-step explanation:

Start off by writing the equation out as it is given:

[tex]f(x)=4(x+6)^{2} +5[/tex]

Then, get handle to exponent and distribution of the 4 outside the parenthesis:

[tex]f(x)=4(x^{2} +36)+5\\f(x)=4x^{2} +144+5[/tex]

Finally, combine any like terms:

[tex]f(x)=4x^{2} +149[/tex]

Answer:

x=1.94

x = - 1.94

Step-by-step explanation:

8x² + 5 = 35

Subtract 5 from each side

8x² + 5-5 = 35-5

8x²  = 30

Divide each side by 8

8x² /8 = 30/8

x² = 15/4

Take the square root of each side

sqrt( x²) = ±sqrt(15/4)

x = ±sqrt(15/4)

x=1.93649

x = - 1.93649

To the nearest hundredth

x=1.94

x = - 1.94

Answer:

1.94

Step-by-step explanation:

[tex]8x^2+5=35\\8x^2=30 \\x^2=30/8\\x^2=3.75\\\sqrt{3.75} \\[/tex]

≈ ±1.94

Answer:

  C. (2, 6] hour jobs cost $100

Step-by-step explanation:

Let's consider each of these statements in view of the graph:

Answer:

(0,8) first option     and  50,100 or 200 in the second option

Step-by-step explanation:

Answer: The lines are parallel.

The slopes are equal.

The y-intercepts are different.

The system has no solution.

Step-by-step explanation:

For  a pair of equations: [tex]a_1x+b_1y=c_1\\\\a_2x+b_2y=c_2[/tex]

They coincide if [tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}[/tex]

They are parallel if [tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}[/tex]

They intersect if [tex]\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}[/tex]

Given equations: [tex]8 x + 10 y = 30\\ 12 x + 15 y = 60[/tex]

Here,

[tex]\dfrac{a_1}{a_2}=\dfrac{8}{12}=\dfrac{2}{3}\\\\ \dfrac{b_1}{b_2}=\dfrac{10}{15}=\dfrac{2}{3}\\\\ \dfrac{c_1}{c_2}=\dfrac{30}{60}=\dfrac{1}{2}[/tex]

⇒[tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}[/tex]  

Hence, The lines are parallel.

It has no solution. [parallel lines have no solution]

Write 8 x + 10 y = 30 in the form of y= mx+c, where m is slope and c is the y-intercept.

[tex]y=-\dfrac{8}{10}x+\dfrac{30}{10}\Rightarrow\ y=-0.8x+3[/tex]

i.e. slope of 8 x + 10 y = 30 is -0.8 and y-intercept =3

Write  12 x + 15 y = 60 in the form of y= mx+c, where m is slope

[tex]y=-\dfrac{12}{15}x+\dfrac{60}{15}\Rightarrow\ y=-0.8x+4[/tex]

i.e. slope of  12 x + 15 y = 60 is -0.8 and y-intercept =4

i.e. The slopes are equal but y-intercepts are different.

Answer: The lines are parallel.

The slopes are equal.

The y-intercepts are different.

The system has no solution.

Step-by-step explanation:

For  a pair of equations:  

They coincide if  

They are parallel if  

They intersect if  

Given equations:  

Here,

⇒  

Hence, The lines are parallel.

It has no solution. [parallel lines have no solution]

Write 8 x + 10 y = 30 in the form of y= mx+c, where m is slope and c is the y-intercept.

i.e. slope of 8 x + 10 y = 30 is -0.8 and y-intercept =3

Write  12 x + 15 y = 60 in the form of y= mx+c, where m is slope

i.e. slope of  12 x + 15 y = 60 is -0.8 and y-intercept =4

i.e. The slopes are equal but y-intercepts are different.

Answer:

A 95​% confidence interval for the proportion of students who eat cauliflower on​ Jane's campus is [0.012, 0.270].

Step-by-step explanation:

We are given that Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 ​students, she finds 2 who eat cauliflower.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

                              P.Q.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of students who eat cauliflower

           n = sample of students

           p = population proportion of students who eat cauliflower

Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                   of significance are -1.96 & 1.96}  

P(-1.96 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 1.96) = 0.95

P( [tex]-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95

P( [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95

Now, in Agresti and​ Coull's method; the sample size and the sample proportion is calculated as;

[tex]n = n + Z^{2}__(\frac{_\alpha}{2})[/tex]

n = [tex]24 + 1.96^{2}[/tex] = 27.842

[tex]\hat p = \frac{x+\frac{Z^{2}__(\frac{\alpha}{2}_) }{2} }{n}[/tex] = [tex]\hat p = \frac{2+\frac{1.96^{2} }{2} }{27.842}[/tex] = 0.141

95% confidence interval for p = [ [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ]

 = [ [tex]0.141 -1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }[/tex] , [tex]0.141 +1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }[/tex] ]

 = [0.012, 0.270]

Therefore, a 95​% confidence interval for the proportion of students who eat cauliflower on​ Jane's campus [0.012, 0.270].

The interpretation of the above confidence interval is that we are 95​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between 0.012 and 0.270.

Answer:

both the equation and it's inverse are functions

Answer:  3.2 yd

Step-by-step explanation:

Notice that TWV is a right triangle.  

Segment TU is not needed to answer this question.

∠V = 32°, opposite side (TW) is unknown, hypotenuse (TV) = 6

[tex]\sin \theta=\dfrac{opposite}{hypotenuse}\\\\\\\sin 32=\dfrac{\overline{TW}}{6}\\\\\\6\sin 32=\overline{TW}\\\\\\\large\boxed{3.2=\overline{TW}}[/tex]

Answer:

a. 24

b.600

c.36

d. No

Step-by-step explanation:

a.You know the approximate number of gallons is about 24 gallons because each tank holds twelve and your family used 2 of them.

b. You know you drove about 600 miles. This is because you used 24 gallons And each gallon should get you 25 miles. multiply The 2 together to get 600 miles. Or you could set a thing like 1/25=24/x and solve for x.

c. It cost 36 dollars because each gallon is 1.5 and you used 24 gallons so mul the two together to get 36

d. First find the amount of gallons used by dividing 350 by 25 to get 14. Then multiply 14 by 1.5 to get 21. 21 is greater than 20 so you don’t have enough money.

Answer:

12 gallons per hour

Step-by-step explanation:

Given the following :

Start time of flight = 9:30 a.m

Arrival time of flight = 1:45p.m

Gallons of gas used during duration of flight = 51 gallons

Number of hours spent during flight:

Arrival time - start time

1:45 pm - 9:30 am = 4hours and 15minutes

4hours 15minutes = 4.25hours

If 4.25hours requires 51 gallons of gas;

Then 1 hour will require ( 51 / 4.25)gallons

= 51 / 4.25

= 12 gallons

Answer:

[tex]\dfrac{1}{x^{48}y^{36}z^{6}}[/tex]

Step-by-step explanation:

[tex] (\dfrac{(x^2y^3)^{-2}}{(x^6y^3z)^{2}})^3 = [/tex]

[tex] = (\dfrac{1}{(x^6y^3z)^{2}(x^2y^3)^{2}})^3 [/tex]

[tex] = (\dfrac{1}{x^{12}y^6z^{2}x^4y^6})^3 [/tex]

[tex]= (\dfrac{1}{x^{16}y^{12}z^{2}})^3[/tex]

[tex]= \dfrac{1}{x^{48}y^{36}z^{6}}[/tex]

Answer:

[tex]\displaystyle \frac{1}{x^{48}y^{36}z^6}[/tex]

Step-by-step explanation:

[tex]\displaystyle[\frac{(x^2 y^3)^{-2}}{(x^6 y^3 z)^2 } ]^3[/tex]

[tex]\displaystyle \frac{(x^2 y^3)^{-6}}{(x^6 y^3 z)^6 }[/tex]

[tex]\displaystyle \frac{(x^{-12} y^{-18})}{(x^{36} y^{18}z^6 ) }[/tex]

[tex]\displaystyle \frac{x^{-48} y^{-36}}{z^6 }[/tex]

[tex]\displaystyle \frac{1}{x^{48}y^{36}z^6}[/tex]

Answer:

z = 44

Step-by-step explanation:

26/z = 13/22

Using cross products

26 * 22 = 13*z

Divide each side by 13

26/13 * 22 = 13z/13

2 *22 =z

44 =z

Answer:

Z=44

To solve this proportion, you have to isolate z. You have to do what you do on one side of the equal sign to the other as well.

Answer:

(C)[tex]x^2+4x-2-(-x^2+2x-4)=2x^2+2x+2[/tex]

Step-by-step explanation:

The expression represented by the upper tiles is: [tex]x^2+4x-2[/tex]

The expression represented by the lower tiles is: [tex]x^2-2x+4[/tex]

Adding the two

[tex]x^2+4x-2+(x^2-2x+4)=2x^2+2x+2[/tex]

Writing it as a difference, we have:

[tex]x^2+4x-2-(-x^2+2x-4)=2x^2+2x+2[/tex]

The correct option is C.

Answer:

yeah, what newton said :]

Answer:

I believe the unit price would be 2.39 per unit

Step-by-step explanation:

Answer:

first option

Step-by-step explanation:

Given

[tex]\frac{15}{x}[/tex] + 6 = [tex]\frac{9}{x^2}[/tex]

Multiply through by x² to clear the fractions

15x + 6x² = 9 ( subtract 9 from both sides )

6x² + 15x - 9 = 0 ( divide through by 3 )

2x² + 5x - 3 = 0 ← in standard form

Consider the factors of the product of the coefficient of x² and the constant term which sum to give the coefficient of the x- term.

product = 2 × - 3 = - 6 and sum = + 5

The factors are + 6 and - 1

Use these factors to slit the x- term

2x² + 6x - x - 3 = 0 ( factor the first/second and third/fourth terms )

2x(x + 3) - 1(x + 3) = 0 ← factor out (x + 3) from each term

(x + 3)(2x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

2x - 1 = 0 ⇒ 2x = 1 ⇒ x = 0.5

Solution set is { - 3, 0.5 }

Answer:

Ram therefore has more amount in his account at the end of the month, and the balances in their bank accounts at the end of the month are arranged in ascending order, i.e. from the smallest to the largest, as follows:

Rahul – Debits AED 200; Rohit – Credits AED 200; and Ram – Credits AED 500.

Step-by-step explanation:

In banking and finance, a credit transaction on a bank account indicates that an additional amount of money has been added to the bank account and the balance has increased. This gives a positive balance in the account

On the other hand, a debit transaction on a bank account indicates that an amount of money has been deducted or withdrawn from the bank account and the balance has therefore reduced. This gives a negative balance in the account.

Based on the above, we have:

a. Ram – Credits AED 500 on 12th May 2020

Since there is no any other credit or debit transaction during the month, this implies that Ram still has Credits AED 500 in his account at the end of the month.

The Credits AED 500 indicates that Ram has a positive balance of AED 500 in his account at the end of the month.

b. Rahul – Debits AED 700 on 12th May 2020 and Credits AED 500 on 15th May 2020.

The balance in the account of Rahul gives Debits of AED 200 as follows:

Debits AED 700 - Credits AED 500 = Debits AED 200

The Debits AED 200 indicates that Rahul has a negative balance of AED 200 in his account at the end of the month.

c. Rohit – Credits AED 700 on 12th May 2002 and Debits AED 500 on 15th May 2020.

The balance in the account of Rohit gives Credits of AED 200 as follows:

Credits AED 700 - Dedits AED 500 = Credits AED 200

The Credits AED 200 indicates that Rohit has a positive balance of AED 200 in his account at the end of the month.

Conclusion

Arrangement of numbers or amounts of money in ascending order implies that they are arranged from the smallest to the largest number or amount.

Since Credits implies positive amount and Debits implies negative amount, Ram therefore has more amount in his account at the end of the month, and the balances in their bank accounts at the end of the month are arranged in ascending order, i.e. from the smallest to the largest, as follows:

Rahul – Debits AED 200; Rohit – Credits AED 200; and Ram – Credits AED 500.

Answer with explanation:

After dilation about the origin(0,0) with the scale factor of 'k" , the image of the original point (x,y) becomes (kx,ky)

From the given graph, the coordinates of point C = (0,6)  [Since it lies on y-axis , the x-coordinate is zero]

After a dilation about the origin(0,0) with the scale factor of [tex]\dfrac{1}{2}[/tex], the new point will be [tex](\dfrac{1}{2}\times0,\dfrac{1}{2}\times6)=(0,3)[/tex]

Now plot this point on y-axis at y=3 as given in the attachment.