After getting a job at the ice cream parlor, Mitzi put $20 in
her savings account every week. If she had $90 in the
account before she began working, which inequality could
be used to determine how many weeks w it would take to
have at least $350 in her account?
A 90w + 20 > 350
B 20w +90 > 350
C 20w – 90 < 350
D w/20<500-90

< 500 - 90
20

Answer:

it is i think :(

Step-by-step explanation:

Answer:

-9

Step-by-step explanation:

(-4-2*-1/2)3

(-4+1)3

-3*3

-9

Answer:

-27

Step-by-step explanation:

Hi there!

[tex](a-2b)^3[/tex]

Plug in the given information:

[tex](-4-2(\displaystyle -\frac{1}{2} ))^3\\\\(-4-(\displaystyle -\frac{2}{2} ))^3\\\\(-4+1)^3\\\\(-3)^3\\\\-27[/tex]

I hope this helps!

Answer:

?

Step-by-step explanation:

An ogive graph serves as a graphical representation of the cumulative relative frequency distribution for quantitative variables. In other words, these graphs plot the percentile on the y-axis and the quantitative variable on the x-axis. An Ogive is connected to a point on the x-axis, that represents the actual upper limit of the last class, i.e.,( 80.5, 0) Take x-axis, 1 cm = 10 marks. Y-axis = 1 cm – 10 c.f. There are two types of ogives : Less than ogive : Plot the points with the upper limits of the class as abscissae and the corresponding less than cumulative frequencies as ordinates. Ogives are useful for determining the median, percentiles and five number summary of data. Remember that the median is simply the value in the middle when we order the data. A quartile is simply a quarter of the way from the beginning or the end of an ordered data set.

I know this is a lot but this will help you understand.

Answer:

B

Step-by-step explanation

TIENES QUE SIEMPRE FIJARTE EL DE ABAJO Y DESPUES EL DE ARRIBA

EN PLAN LA LINEA QUE PASA POR EL EJE

Answer:1 second.

1/60 of an hour is a minute so 1/60 of a minute is a second

Answer:

[tex]\large\boxed{\tt-\frac{3}{4} x}[/tex]

Step-by-step explanation:

[tex]\textsf{We are asked to identify the missing number for our equation.}[/tex]

[tex]\textsf{First, we should translate our \underline{word form} equation into \underline{numerical form}.}[/tex]

[tex]\tt \underset{-4}-4 \ \underset{\times}{times} \ \underset{?}{what} \ \underset{=}{equals} \ \underset{3x}{3x}?}[/tex]

[tex]\tt -4 \times \ ? = 3x[/tex]

[tex]\large\underline{\textsf{Identifying Ways of Solving;}}[/tex]

[tex]\textsf{We know that the unknown value is multiplied by -4 to equal 3x.}[/tex]

[tex]\textsf{-4 will equal the quotient of 3x and the unknown number.}[/tex]

[tex]\textsf{Hence, the unknown number will equal the quotient of 3x and 4.}[/tex]

[tex]\large\underline{\textsf{Solving For the Unknown Number;}}[/tex]

[tex]\tt \frac{3x}{-4} = Unknown \ Number[/tex]

[tex]\underline{\textsf{Multiply the Numerator and Denominator by the Reciprocal of -4;}}[/tex]

[tex]\textsf{The Reciprocal is a fraction where the numerator and denominator flip.}[/tex]

[tex]\tt Reciprocal \ of \ -4 \ is; -\frac{4}{1} \rightarrow \boxed{-\frac{1}{4}}[/tex]

[tex]\mathtt{ \frac{-\frac{1}{4} \times3x}{-\frac{1}{4} \times -4} \rightarrow \frac{\frac{-3}{4}x }{1} \rightarrow \large\boxed{\tt-\frac{3}{4} x}}[/tex]

Answer:

Define:

Domain is the set of values that you can input in a function

Range is the set of values that corresponds to the output of a function, the set of values corresponding to its results

The vertical line test is a simple test you can do with a graph to know if this graph correspond to a function or not: you have to draw a vertical line, parallel to the y-axis and if that line touch the graph in just one point, it is a function

What is the domain and range of each function?

Domain = {3, 5, 7, 8, 11}

Range = {6, 7, 9, 14}

Is each relation a function?

Yes

Answer:

my butt

Step-by-step explanation:

why well because when were dinosaurs created in 1957 and when they were alive they drove cars to school to learn about extra chromosomes and and then the meatore hit and created my butt.

I hope that helps if you need anymore help let me know :)

Mark as brainliest :)

Answer:

D = -6

Step-by-step explanation:

2d= 6 + 3d

Subtract 3d from each side

-d =6

divide by a -1

d = -6

Answer:

d=(-6)

Step-by-step explanation:

2×(-6)=(-12)=6+(3×(-6))

Answer: a

Step-by-step explanation:

Answer:

A robot can complete tasks in

hour. Each task takes the same amount of time.

a. How long does it take the robot to complete one task?

It takes one hour to complete one task.

b. How many tasks can the robot complete in one hour?

It complete only one task in one hour.

Answer:

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

Explanation:

The first pic shows the line graphed, while the second pic shows how to turn the equation into proper slope-intercept form.

The only steps the second pic is missing however, is flipping the sides around to be better understood (optional), and more importantly subtracting 2 on both sides in order to move it to the other side, to get 2y = x - 2. And dividing the 2 on both sides to get y by its self, to get:

y = (x-2)/2 , or y = 1/2 x - 2/2 , which if you notice 2/2 makes it equal to one whole so your answer is y = 1/2 x - 1 .

If you like this answer can you please mark me brainliest? :)

Answer:

Banks play an important role in the economy by offering a service for people wishing to save. Banks also play an important role in offering finance to businesses that wish to invest and expand. These loans and business investment are important for enabling economic growth.

The financial services sector is the primary driver of a nation's economy. It provides the free flow of capital and liquidity in the marketplace. When the sector is strong, the economy grows, and companies in this industry are better able to manage risk.

Step-by-step explanation:

Answer:

C. 11 customers

Step-by-step explanation:

The pattern is that it rises or goes up by 2 in the y axis, it runs or goes to the right once in the x-axis.

Answer:

3y^5(2y^3+3)=−6y^8−9y^5

−2c^6(5c^3+6)=−10c^9−12c^6

are the correct equations.

Step-by-step explanation:

If you simply all these equations you will get following equations :

when bases are same and are in multiplication we add the exponents.

−4b^4(5b^2+6)=−20b^6−24b^4

−5x^3(3x^3+2)=−15x^6−10x^3

−3y^5(2y^3+3)=−6y^8−9y^5

−2c^6(5c^3+6)=−10c^9−12c^6

so equation 3 and 4 are correct.

8 boxes

Step-by-step explanation:

Start with a proportion

32/72

reduce

4/9

so now

4 jars of raspberry and 9 strawberry times 8 for the total number

so

8 boxes

Answer:

Any expression that sums up to 96 is equivalent to 12+84.

Step-by-step explanation:

Since, you haven't provided any options.

12+84=96

Any expression that sums up to 96 is equivalent to 12+84.

9514 1404 393

Answer:

Step-by-step explanation:

Jarvis is paid for 1 1/2 times the number of hours he worked on Saturday. So, for the 5 hours he worked, he is paid for ...

  5 × 1 1/2 = 7 1/2 hours . . . . paid on Saturday

Jarvis is paid for 1 3/4 times the number of hours he worked on Sunday, So, for the 2 hours he worked, he is paid for ...

  2 × 1 3/4 = 3 1/2 hours . . . . paid on Sunday

The total of hours for which Jarvis is paid is ...

  7 1/2 + 3 1/2 = 11 hours . . . . paid for the weekend

At $9 per hour, the amount Jarvis is paid for the weekend is ...

  $9 × 11 = $99.00 . . . . weekend pay

Answer:

it will rotate 7.5 degrees

Step-by-step explanation:

Answer:

Suppose the universal set is U = all whole numbers from 1 to 9. If A = {1, 2, 4}, then Ac = {3, 5, 6, 7, 8, 9}. As we saw earlier with the expression Ac ...

Let U be the set of whole numbers from 5-10 inclusive. Set A={5, 7, 9 ...

Answer:

I wrote the answers in the screenshot

Answer:

80 I believe

Step-by-step explanation:

(1)

(a) Use the fact that [tex]\sqrt{x^2} = |x|[/tex] for all [tex]x[/tex]. Since [tex]x\to+\infty[/tex], we have [tex]x>0[/tex] and [tex]|x| = x[/tex].

[tex]\displaystyle \lim_{x\to\infty} \frac{\sqrt{9x^2 - 2}}{x + 4} = \lim_{x\to\infty} \frac{\sqrt{x^2} \sqrt{9 - \frac2{x^2}}}{x + 4} \\\\= \lim_{x\to\infty} \frac{ x \sqrt{9 - \frac2{x^2}}}{x+4} \\\\= \lim_{x\to\infty} \frac{\sqrt{9 - \frac2{x^2}}}{1 + \frac4x} \\\\= \frac{\sqrt9}1 = \boxed{3}[/tex]

(b) This time [tex]x\to-\infty[/tex], so [tex]x < 0[/tex] and [tex]|x| = -x[/tex].

[tex]\displaystyle \lim_{x\to\infty} \frac{\sqrt{9x^2 - 2}}{x + 4} = \lim_{x\to\infty} \frac{\sqrt{x^2} \sqrt{9 - \frac2{x^2}}}{x + 4} \\\\= \lim_{x\to\infty} \frac{ \boxed{-x} \sqrt{9 - \frac2{x^2}}}{x+4} \\\\= (-1) \times \lim_{x\to\infty} \frac{\sqrt{9 - \frac2{x^2}}}{1 + \frac4x} \\\\= -\frac{\sqrt9}1 = \boxed{-3}[/tex]

(c) We immediately have

[tex]\displaystyle \lim_{x\to\infty} (x - \sqrt x) = \boxed{\infty}[/tex]

since [tex]x > \sqrt x[/tex] for all [tex]x > 1[/tex].

(d) Introduce a difference of squares by factoring in the limand's conjugate. The rest mirrors what we did in (a)/(b).

[tex]\displaystyle \lim_{x\to\infty} \left(\sqrt{x^2 + 12x} - x\right) = \lim_{x\to\infty} \frac{\left(\sqrt{x^2+12x} - x\right) \left(\sqrt{x^2+12x} + x\right)}{\sqrt{x^2 + 12x} + x} \\\\ = \lim_{x\to\infty} \frac{\left(\sqrt{x^2+12x}\right)^2 - x^2}{\sqrt{x^2+12x}+x} \\\\ = \lim_{x\to\infty} \frac{12x}{\sqrt{x^2+12x}+x} \\\\ = \lim_{x\to\infty} \frac{12}{\sqrt{1 + \frac{12}x} + 1} = \frac{12}{\sqrt1 + 1} = \boxed{6}[/tex]

(e) Divide through by the highest-degree exponential term.

[tex]\displaystyle \lim_{x\to\infty} \frac{12e^{2x} - 3e^{3x}}{2e^{2x} + 4e^{3x}} = \lim_{x\to\infty} \frac{12e^{-x} - 3}{2e^{-x} + 4} = \frac{0 - 3}{0 + 4} = \boxed{-\frac34}[/tex]

(2) By definition of the derivative, we have

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h) - f(x)}h[/tex]

For [tex]f(x) = \sqrt{x^2+1}[/tex], the limit becomes

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{\sqrt{(x+h)^2+1} - \sqrt{x^2+1}}h[/tex]

Factor in the conjugate.

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{\left(\sqrt{(x+h)^2+1} - \sqrt{x^2+1}\right) \left(\sqrt{(x+h)^2+1} + \sqrt{x^2+1}\right)}{h \left(\sqrt{(x+h)^2+1} + \sqrt{x^2+1}\right)}[/tex]

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{\left(\sqrt{(x+h)^2+1}\right)^2 - \left(\sqrt{x^2+1}\right)^2}{h \left(\sqrt{(x+h)^2+1} + \sqrt{x^2+1}\right)}[/tex]

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{\bigg((x+h)^2+1\bigg) - (x^2+1)}{h \left(\sqrt{(x+h)^2+1} + \sqrt{x^2+1}\right)}[/tex]

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{2xh + h^2}{h \left(\sqrt{(x+h)^2+1} + \sqrt{x^2+1}\right)}[/tex]

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{2x + h}{\sqrt{(x+h)^2+1} + \sqrt{x^2+1}}[/tex]

[tex]\implies \boxed{f'(x) = \displaystyle \lim_{h\to0} \frac{x}{\sqrt{x^2+1}}}[/tex]

(3) The tangent line to

[tex]y = \frac1{x^2+1}[/tex]

at the point (2, 1/5) has slope equal to the derivative [tex]\frac{dy}{dx}[/tex] when [tex]x = 2[/tex]. Compute the derivative; since [tex]y = \frac1{f(x)^2}[/tex] where [tex]f(x)[/tex] is the function from the previous problem, using the chain rule gives

[tex]y = \dfrac1{f(x)^2} \implies \dfrac{dy}{dx} = -\dfrac{2f'(x)}{f(x)^3} = -\dfrac{2 \times \frac{x}{\sqrt{x^2+1}}}{\left(\sqrt{x^2+1}\right)^3} \\\\ \implies \dfrac{dy}{dx} = -\dfrac{2x}{(x^2+1)^2}[/tex]

The tangent line at (2, 1/5) then has slope

[tex]\dfrac{dy}{dx}\bigg|_{x=2} = -\dfrac{2\times2}{(2^2+1)^2} = -\dfrac4{25}[/tex]

Using the point-slope formula, the equation of the tangent line is

[tex]y - \dfrac15 = -\dfrac4{25} (x - 2) \implies \boxed{y = -\dfrac{4x - 13}{25}}[/tex]

Answer:

7.083

Step-by-step explanation:

well there's the ones place, tenths, hundredths, and thousandths. easy math, unless i'm stoopid and this is wrong :D

Answer:

1/10

Step-by-step explanation:

The default placeholder value for tenths in math is 0.1, or 1/10. the first digit after the decimal point.

y=mx+b

-3=3(3)+b

-3=9+b

-12=b