Steve bought a new car for $22,000 but paid 93% of the list price. How much was the list price

Answer:

Answer:

Step-by-step explanation:

we are essentially calculating:

4(1/4)^0 + 4(1/4)^1 + 4(1/4)^2 + ... + 4(1/4)^9

Formula: [tex]\frac{a(1-r^n)}{1-r}[/tex]

4(1-(1/4)^10)/ (1-1/4) approximates to 5.33

D. 12.5

The equation to find the average rate of change is

[tex] \frac{y2 - y1 }{x2 - x1} [/tex]

To find the average rate of change between points (0,20) and (8,120) you need to plug it into the equation so it'll be

[tex] \frac{120 - 20}{8 - 0} [/tex]

this simplifies to

[tex] \frac{100}{8} [/tex]

which is 12.5

Answer:

[tex]2x^{2}[/tex] - 16x + 14

Step-by-step explanation:

(2x - 4)(x - 6)

= (2x + −4)(x + −6)

= (2x)(x) + (2x)(−6) + (−4)(x) + (−4)(−6)

= [tex]2x^{2}[/tex] − 12x − 4x + 24

= [tex]2x^{2}[/tex] - 16x + 14

The values that are part of the solution are x = -5 and x = -4.5

Number lines are lines that are used to represent the solution to inequalities.

From the given number line with he solution x > -4, this shows that all the solutions must be values that are greater than -4.

The values that are part of the solution are x = -5 and x = -4.5

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Answer:

28 feet

Step-by-step explanation:

Area of a rectangle is :

A = L × W

We are given A and a expression for L, so let's substitute :

448 = (2W-4) × W

Now we expand the left side :

2W²-4w = 448

Subtract 448 from both sides :

2W²-4W-448 = 0

Divide everything by 2 :

W²-2W-224 = 0

Now we factorise :

Find 2 numbers that multiply to give -224 and add to give -2 :

-16 and 14

Rewrite -2W with -16W and +14W :

W² - 16W + 14W -224 = 0

W(W-16)  +14(W-16) = 0

(W+14)(W-16) = 0

W = -14 ,  W = 16

Only take positive value since this is lengths :

W = 16

Now we substitute W into the expression to find L :

L = (2W -4)

L = 2(16) - 4

L = 32 - 4

L = 28

Units will be feet since it is a length

Hope this helped and have a good day

Answer:

390/1000

This is the answer and hope it helps

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \: f(0) = -2[/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

[tex]\qquad \tt \rightarrow \: 2 {x}^{2} - 3y = 6[/tex]

[tex]\qquad \tt \rightarrow \: 3y + 6 = 2 {x}^{2} [/tex]

[tex]\qquad \tt \rightarrow \: 3y = 2 {x}^{2} - 6[/tex]

[tex]\qquad \tt \rightarrow \: y = \cfrac{2 {x}^{2} - 6 }{3} [/tex]

[tex]\qquad \tt \rightarrow \: y = \cfrac{2 {}^{} }{3} {x}^{2} - 2[/tex]

[ here, y can be replaced with f(x) because y is a function of x ]

[tex]\qquad \tt \rightarrow \: f(x) = \cfrac{2 {}^{} }{3} {x}^{2} - 2[/tex]

[tex] \large\textsf{Find f(0):} [/tex]

[tex]\qquad \tt \rightarrow \: f(0) = \cfrac{2 {}^{} }{3} {(0)}^{2} - 2[/tex]

[tex]\qquad \tt \rightarrow \: f(0) = 0 - 2[/tex]

[tex]\qquad \tt \rightarrow \: f(0) = - 2[/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

Answer:

Maybe    9     and 12   ?

Step-by-step explanation:

-6 + 9 = 3  

    then   3 + 9 = 12

           then 12 + 3 = 15      

                 each term is the sum of the two that are before it

Using it's concept, it is found that the domain of the function is given as follows:

d. 0 < c ≤ 15

The domain is the set that contains all possible input values for the function.

He is losing weight consistently, hence c > 0, and the maximum amount is of 15 calories, hence c ≤ 15 and the domain is given by:

d. 0 < c ≤ 15

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first one b 315, second d

Step-by-step explanation:

18*=3*+315

7*+4*=a*

Answer:

The solution is 4 - 9/(2x + 1)

No, the divisor does not evenly divide into the dividend

Step-by-step explanation:

Please see the attached image

Answer:

[tex]7 469.8.56[/tex]

nossa nem sei como é bom saber se fala com a mãe e o pai do mundo e vc não tem mais nada pra comer e não tem como fazer a gente vai ficar muito tempo sem fazer a mãe dela não e vc como tá o trabalho rede de nada em casa e é só pra mim né não é fácil de fazer a mãe dela e a mãe dele tá bem né não é fácil de fazer o que vc quer que Deus quiser vai dar pra mim né e não tem como fazer d id da x não vou poder é o que eu te abençoe e ilumine e ilumine todos os dias entre?01029)

Answer:

2:45

Step-by-step explanation:

1 1/2 hour is 1 hour and 30 minutes.

If we add 30 minutes to 1:15, we get 1:45

if we add 1 hour to 1:45, we get 2:45, which is our answer.

Answer:

Using the formula multiplicity, we find that the equation of the function will be [tex]f(x)=-(x+6)^{3} (x-2)^{4}[/tex]. The graph is in the attachment.

Step-by-step explanation:

Concept: Given that for -6, multiplicity is 3 and for 2 multiplicity is 4.

So, the equation of multiplicity is represented as:

[tex]f(x)=a(x-root)^{mutliplicity}[/tex]

This gives the following function

[tex]f(x)=a(x+6)^{3} (x-2)^{4}[/tex]

The equation has a negative leading coefficient.

This means that, the value of a is less than 0 i.e. a < 0

Assume any value of a (say a = -1), the equation becomes

[tex]f(x)=-(x+6)^{3} (x-2)^{4}[/tex]

The graph is in the attachment.

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Answer:

C

Step-by-step explanation:

trust bro

Answer:

When x = 10, they will both be y = 60

(10, 60)

Step-by-step explanation:

2(10) + 40 = 60

4(10) + 20 = 60

Answer:

x = 10

Step-by-step explanation:

2 (10) +40 = 60

4(10) +20 = 60

Answer: Beans cause Nia to Fart!

Step-by-step explanation:

Mexican food usually contains beans, which beans are a known catalyst of flatulence. Beans induce farts and Nia does so! Farts usually stink, so an air freshener is needed.

Answer:

side ≈ 4.24= [tex]\frac{6}{\sqrt{2} }[/tex]

Step-by-step explanation:

sides a = b

[tex]6^{2} =a^{2} +a^{2}[/tex]

[tex]6^{2} =2a^{2}[/tex]

[tex]a^{2}= \frac{6^{2} }{2} =\frac{36}{2} =18[/tex]

[tex]a=\sqrt{18} =4.24=\sqrt{(9)(2)} =\frac{6}{\sqrt{2} }[/tex]

Hope this helps

Answer:

48 cm²

Explanation:

The figure can be cut into 1 rectangle and 1 triangle.

Area of the figure:

⇒ area of rectangle + area of triangle

⇒ Length × Width  +  1/2 × Base × Height

⇒ 6 × 4  +  1/2 × 6 × 8

⇒ 24  +  24

⇒ 48

Therefore, area of figure is 48 cm².

Answer:

48 cm squared

Step-by-step explanation:

The given figure is composed of a rectangle (L=6 cm and W = 4 cm) and a right angled triangle { Base = (12- 4) = 8 cm, height = 6 cm. }

Area of the two figures is to be calculated separately and then added to get the required result.

Area of the rectangle = l × w

= 6 × 4

= 24 cm²

Area of the right triangle = ½ × base × height

= ½ × 8 × 6

= 24 cm²

Hence, area of the given figure = 24 + 24

= 48 cm squared.

We have y+2 = 0 and x - 2 = 0. The provided function has an x and y-intercept of -2 and +2, respectively. There is no vertical asymptote. Two is the horizontal asymptote.

A diagram depicting the relationship between two or more variables, each measured along with one of a pair of axes at right angles.

The y-intercept of a function is determined by the intersection of its graph with the y-axis. The value of y on the y-axis at which the considered function crosses it is called the y-intercept.

Assume the following equation: y = f (x)

We have x   =0- 2 and y+2 = 0,The x and y intercept of the given function is -2 and +2.

The vertical asymptote is none. The horizontal asymptote is 2.

Hence,we have y+2 = 0 and x - 2 = 0. The provided function has an x and y-intercept of -2 and +2, respectively. There is no vertical asymptote. Two is the horizontal asymptote.

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If the geometric series has first term [tex]a[/tex] and common ratio [tex]r[/tex], then its [tex]N[/tex]-th partial sum is

[tex]\displaystyle S_N = \sum_{n=1}^N ar^{n-1} = a + ar + ar^2 + \cdots + ar^{N-1}[/tex]

Multiply both sides by [tex]r[/tex], then subtract [tex]rS_N[/tex] from [tex]S_N[/tex] to eliminate all the middle terms and solve for [tex]S_N[/tex] :

[tex]rS_N = ar + ar^2 + ar^3 + \cdots + ar^N[/tex]

[tex]\implies (1 - r) S_N = a - ar^N[/tex]

[tex]\implies S_N = \dfrac{a(1-r^N)}{1-r}[/tex]

The [tex]N[/tex]-th partial sum for the series of reciprocal terms (denoted by [tex]S'_N[/tex]) can be computed similarly:

[tex]\displaystyle S'_N = \sum_{n=1}^N \frac1{ar^{N-1}} = \frac1a + \frac1{ar} + \frac1{ar^2} + \cdots + \frac1{ar^{N-1}}[/tex]

[tex]\dfrac{S'_N}r = \dfrac1{ar} + \dfrac1{ar^2} + \dfrac1{ar^3} + \cdots + \dfrac1{ar^N}[/tex]

[tex]\implies \left(1 - \dfrac1r\right) S'_N = \dfrac1a - \dfrac1{ar^N}[/tex]

[tex]\implies S'_N = \dfrac{1 - \frac1{r^N}}{a\left(1 - \frac1r\right)} = \dfrac{r^N - 1}{a(r^N - r^{N-1})} = \dfrac{1 - r^N}{a r^{N-1} (1 - r)}[/tex]

We're given that [tex]a=1[/tex], and the sum of the first [tex]n[/tex] terms of the series is

[tex]S_n = \dfrac{1-r^n}{1-r} = 364[/tex]

and the sum of their reciprocals is

[tex]S'_n = \dfrac{1 - r^n}{r^{n-1}(1 - r)} = \dfrac{364}{243}[/tex]

By substitution,

[tex]\dfrac{1 - r^n}{r^{n-1}(1-r)} = \dfrac{364}{r^{n-1}} = \dfrac{364}{243} \implies r^{n-1} = 243[/tex]

Manipulating the [tex]S_n[/tex] equation gives

[tex]\dfrac{1 - r^n}{1-r} = 364 \implies r (364 - r^{n-1}) = 363[/tex]

so that substituting again yields

[tex]r (364 - 243) = 363 \implies 121r = 363 \implies \boxed{r=3}[/tex]

and it follows that

[tex]r^{n-1} = 243 \implies 3^{n-1} = 3^5 \implies n-1 = 5 \implies \boxed{n=6}[/tex]

The statement about both functions that is true is:

The domain of a function is the set of values of input for which the function is valid.

The range is the dependent variable of a set of values for which the function is defined.

Given that:

f(x) = 4 - x

For function g(x) = -1 ×[tex]\mathbf{\sqrt{x-4}}[/tex]

Therefore, from the given options, the statement about both functions that is true is:

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Answer:

Both f(x) and g(x) include domain values of [4, ∞), and both functions decrease over the interval (4, ∞).

Step-by-step explanation:

I got it right on the test.

The clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

The clock was still accurate by Friday noon. The clock was late by 468 seconds by Monday, 6 pm.

To solve the problem, we must:

Know how many 30-minutes have passed during the time period.

1 day = 24 hours

1 hour = 60 minutes = 2 × (30 minutes)

1 day = 24 hours × 2 × (30 minutes)

1 day = 48 × (30 minutes)

Thus, there are 48, 30-minutes in a day. On Friday, however, we start counting at noon, which is half of the day. Moreover, on Monday, the mark is only up to 6 pm, which is three-fourths of the day.

Friday = 48 × [tex]\frac{1}{2}[/tex] = 24

Saturday = 48

Sunday = 48

Monday = 48 × [tex]\frac{3}{4}[/tex] = 36

TOTAL = 24 + 48 + 48 + 36 = 156

Therefore, the total number of 30-minutes that have passed is 156. There were 156, 30-minutes that passed during the time period.

Divide the number of total seconds late by the number of 30-minutes passed.

That is, the number of total seconds late= 468 seconds ÷ 156

= 3 seconds  

Therefore, the clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

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Answer:

                                  [tex]f^{\prime}\left(x\right)\ =\ -\frac{65}{2}x^{\frac{11}{2}}\ +\frac{49}{2}x^{-\frac{9}{2}}[/tex]

or

                                  [tex]f^{\prime}\left(x\right)\ =\ -32.5x^{5.5}\ +\ 24.5x^{-4.5}[/tex]

Step-by-step explanation:

Rather than solving this question in a more complex method by directly using the product rule and quotient rule, it can first be considered to perform some algebraic manipulation (index laws) to simplify the expression before taking the derivative.

                                       [tex]\begin{large}\begin{array}{l}f\left(x\right)\ =\ -5x^6\ \sqrt{x}\ +\ \frac{-7}{x^3\ \sqrt{x}}\\\\f\left(x\right)\ =\ -5x^6\cdot x^{\frac{1}{2}}\ +\ \frac{-7}{x^3\cdot x^{\frac{1}{2}}}\\\\f\left(x\right)\ =\ -5x^{6\ +\ \frac{1}{2}}\ +\ \frac{-7}{x^{3\ +\ \frac{1}{2}}}\\\\f\left(x\right)\ =\ -5x^{\frac{13}{2}}\ +\ \frac{-7}{x^{\frac{7}{2}}}\\\\f\left(x\right)\ =\ -5x^{\frac{13}{2}}\ -7x^{-\frac{7}{2}}\end{array}[/tex]

Now, the derivative of the function can be calculated simply by only using the power rule, which yields

                   [tex]\begin{large}\begin{array}{l}f\left(x\right)\ =\ -5x^{\frac{13}{2}}\ -7x^{-\frac{7}{2}}\\\\f^{\prime}\left(x\right)\ =\ \left(-5\right)\left(\frac{13}{2}\right)\left(x^{\frac{13}{2}\ -\ 1}\right)\ -\ \left(7\right)\left(-\frac{7}{2}\right)\left(x^{-\frac{7}{2}\ -\ 1}\right)\\\\f^{\prime}\left(x\right)\ =\ -\frac{65}{2}x^{\frac{11}{2}}\ +\frac{49}{2}x^{-\frac{9}{2}}\\\\f^{\prime}\left(x\right)\ =\ -32.5x^{5.5}\ +\ 24.5x^{-4.5}\end{array}\\\end{large}[/tex]

Answer:

B

Step-by-step explanation:

4, 16, 64 should be the correct order

Answer:

11/200

Step-by-step explanation:

Convert the percentage to a fraction by placing the expression over

100

Percentage means 'out of 100

5.5

100

Convert the decimal number to a fraction by shifting the decimal point in both the numerator and denominator. Since there is

1

number to the right of the decimal point, move the decimal point

1

place to the right.

55

1000

Cancel the common factor of

55

and

1000

11/200

Eric spend 60 minutes for basketball and 35 minutes for volleyball.

An equation that has the highest degree of 1 is known as a linear equation. This means that no variable in a linear equation has an exponent more than 1. The graph of a linear equation always forms a straight line.

Here, let Eric plays basketball for x minutes.

          and volleyball for y minutes.

Total time spend on playing basketball and volleyball everyday = 95 minutes.

Duration of Basketball 25 minutes more than volleyball

Now, according to question,

          x + y = 95        ..........(i)

          x - 25 = y          ...........(ii)

On solving equation (i) and (ii), we get

        x + x - 25 = 95

           2x = 95 + 25

           2x = 120

            x = 120 / 2

            x = 60

put value of x in equation (ii), we get

        60 - 25 = y

          y = 35

Yes, it is possible and clearly described in above calculation.

Thus, Eric spend 60 minutes for basketball and 35 minutes for volleyball.

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