Sarah's craft project uses pieces of yarn that are 1/8 yard long. She has a piece of yarn that is 3 yards long. How many 1/8 yard pieces can she cut and still have 1 1/4 yards left?

Answer:

f(x) = (x - 3)²

Step-by-step explanation:

If the function given below is g(x) = x²

Then we have to find the function f(x) which is same as g(x) but shifted 3 units left.

As we know in the given equation f(x) = (a + x)²+ k, x defines the shift of the function and and k defines the shift of a graph on y-axis upwards or downwards.

Here the graph is shifted left so the value of x will be (-3) which defines the shifting of 3 left on x-axis.

Therefore the new function will be f(x) = (x - 3)²

Answer:

F(x)=(x+3)^2

Step-by-step explanation:

The results of the expression are 8 2/7 ; 16.286 and 16 respectively.

Using the expression given :

Now we divide:

A.)

456 / 28 = 16 remainder 8

Now we have :

8 8/28

= 8 2/7

B.)

As a decimal :

456 / 28

= 16.286

C.)

Rounded to the nearest whole number:

456 / 28

= 16.286

≈ 16

Final answer:

To divide 456 by 28: as a mixed number is 16 2/7, as a decimal rounded to two places is 16.29, and to the nearest whole number is 16.

Explanation:

To divide 456 by 28, we start by performing the division to find the quotient and the remainder. Firstly, 456 divided by 28 equals 16 with a remainder of 8.

Answering the three parts:

The information provided seems to be examples of rounding and significant figures from various arithmetic operations but does not directly relate to the division of 456 by 28, which is our primary focus. Please note that while rounding to the nearest whole number or to two decimal places, it's important to correctly evaluate the quotient that the specific calculation yields.

In this mathematical problem involving weight reduction over time, Keith's dog has to lose 15 pounds to reach a healthy weight. If the dog can lose an average of 1.25 pounds per week, it would take the dog 12 weeks to achieve the desired weight.

The subject of the question is Mathematics, specifically a problem concerning weight loss over time, which is a real-world application of arithmetic and rate problems. In the given scenario, Keith's dog weighs 90 pounds, and the desired healthy weight is less than 75 pounds. Therefore, the dog needs to lose a total of 15 pounds. Given the dog can lose an average of 1.25 pounds per week, we can calculate the number of weeks it would take for the dog to reach a healthy weight.

Step 1: Identify the total amount of weight to be lost, which is the difference between the current weight and desired weight. This is calculated as 90 pounds - 75 pounds = 15 pounds.

Step 2: Divide the total weight to be lost by the average weight loss per week. This is calculated as 15 pounds ÷ 1.25 pounds per week = 12 weeks.

Therefore, it would take the dog 12 weeks to reach a healthy weight if he loses an average of 1.25 pounds per week.

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Answer:

the answer is b.63

Step-by-step explanation:

Hello, I can help you with this

Step 1

to solve this you must divide the money Mr Smith has by the price of a stock per share, it is

Money Mr smith has = $200

The price of a stock is $3.17

[tex]\frac{200 usd}{3.17\frac{usd}{share} }\\ =63.09\ shares\\\\[/tex]

he cand by  whole shares, so he can buy exactly 63 shares

so, the answer is b.63

I really hope it helps, Have a great day.

Final answer:

To express[tex]\(\sqrt{x^3}\)[/tex] as a fractional exponent, it is rewritten as[tex]\(x^{\frac{3}{2}}\),[/tex] correlating the square root operation to the power of \[tex](\frac{1}{2}\)[/tex]and multiplying by the exponent 3.

Explanation:

To change the radical[tex]\(\sqrt{x^3}\)[/tex]to an algebraic expression with fractional exponents, we can use the property that any root can be expressed as a power with a fractional exponent. The square root of a variable to some power can be represented by raising the variable to the power of half the original exponent.

In this case,[tex]\(\sqrt{x^3}\)[/tex]can be written as [tex]\(x^{\frac{3}{2}}\).[/tex] This is because the square root operation is equivalent to raising to the power of[tex]\(\frac{1}{2}\),[/tex] and when we multiply the exponents [tex]\(3 \times \frac{1}{2}\),[/tex] we get[tex]\(\frac{3}{2}\).[/tex]

Answer:

Step-by-step explanation:

A straight line can either have 1 y-intercept or infinitely many y intercepts. The only case in which there are infinitely many y-intercepts is when the line represents the y-axis itself. On y-axis, x is always 0. So the line that represents the y-axis is the line x=0. However, it is not a linear equation. In short, only the line representing y-axis has more than 1 y-intercepts. There cannot be such a linear equation(function).

A linear equation cannot have more than one y-intercept. Having multiple y-intercepts would violate the definition of a function as each input (x-value) should only be associated with one output (y-value). Therefore, a linear equation with multiple y-intercepts suggests an error.

In the field of mathematics, specifically in algebra and graphing, a linear equation cannot have more than one y-intercept. The y-intercept of a linear equation is the point at which the line crosses the y-axis on a coordinate plane. Linear equations are represented by the formula y = mx + b, where 'm' is the slope of the line and 'b' is the y-intercept. If an equation were to have more than one y-intercept, it would violate the definition of a function, which states that each input (or x-value) is only associated with one output (or y-value). Hence, if a linear equation was portrayed as having more than one y-intercept, it would imply an error in the equation or in the graph itself.

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Answer:  They work for 2.5 hours to gift wrap  65 boxes.

Step-by-step explanation:

Since we have given that

Number of boxes Jack can wrap = 28

Time taken by him = 2 hours

Number of boxes Walter can wrap = 36

Time taken by him = 3 hours

Work done by Jack in  1 hour is given by

[tex]\frac{2}{28}\\\\=\frac{1}{14}[/tex]

Work done by Walter in 1 hour is given by

[tex]\frac{3}{36}\\\\=\frac{1}{12}[/tex]

So, Work done by both Jack and Walter in 1 hour is given by

L.C.M. of [tex](\frac{1}{14},\frac{1}{12})[/tex] is [tex]=\frac{l.c.m.\ of \ numerator}{h.c.f\ of\ denominator}=\frac{1}{2}[/tex]

Number of units done by Jack is given by

[tex]\frac{\frac{1}{2}}{\frac{1}{14}}\\\\=\frac{14}{2}\\\\=7\ units[/tex]

Number of units done by Walter is given by

[tex]\frac{\frac{1}{2}}{\frac{1}{12}}\\\\=\frac{12}{2}\\\\=6\ units[/tex]

So, Let the number of hours be 't' .

[tex]7t+6t=\frac{1}{2}\times 65\\\\13t=\frac{1}{2}\times 65\\\\t=\frac{1}{2}\times \frac{65}{13}\\\\t=\frac{5}{2}\\\\t=2.5\ hours[/tex]

So, They work for 2.5 hours to gift wrap  65 boxes.

The Answer is B.

I took the unit review, as well as did the math myself,

Answer:

Average velocity is 32 miles/hr.

Step-by-step explanation:

Given that a particle moves on a line away from its initial position so that after t hours it is  [tex]s=6t^2+2t[/tex] miles from its initial position.

We have to find the average velocity of the particle over the interval [1, 4].

As, average velocity is the change is position over the change in time.

[tex]s(4)=6(4)^2+2(4)=104[/tex]

[tex]s(1)=6(1)^2+2(1)=8[/tex]

∴ [tex]\text{Average Velocity=}\frac{s(4)-s(1)}{4-1}[/tex]

                            =[tex]\frac{104-8}{3}=\frac{96}{3}=32miles/hr[/tex]

Hence, average velocity is 32 miles/hr.

The average velocity of the particle is [tex]\boxed{{\mathbf{21 units}}}[/tex] .

Further explanation:  

Velocity is the speed of an object in a given direction. Velocity is the vector quantity.

The average velocity can be calculated as,

[tex]{\text{average velocity}}=\frac{{{\text{distance travelled}}}}{{{\text{time taken}}}}[/tex]

Given:

The position of the particle after [tex]t[/tex]  hours is [tex]s\left(t\right)=4{t^2}+t[/tex] . The given interval is [tex]\left[{1,4}\right][/tex] .

Step by step explanation:

Step 1:  

The position of the particle after [tex]t[/tex]  hours is [tex]s\left(t\right)=4{t^2}+t[/tex]  

First we need to find the distance travelled in the interval of [tex]\left[{1,4}\right][/tex] .

The distance travelled by the particle at [tex]t=1[/tex]  is as follows,

[tex]\begin{gathered}s\left(t\right)=4{\left(t\right)^2}+t\hfill\\s\left(1\right)=4{\left(1\right)^2}+1\hfill\\s\left( 1 \right)=5\hfill\\\end{gathered}[/tex]

The distance travelled by the particle at [tex]t=4[/tex]  is as follows,

[tex]\begin{gathered}s\left(t\right)=4{\left(t\right)^2}+t\hfill\\s\left(4\right)=4{\left(4\right)^2}+1\hfill\\s\left( 1 \right)=68\hfill\\\end{gathered}[/tex]

Now find the distance travelled by the particle in the interval of [tex]\left[{1,4}\right][/tex]  .

[tex]\begin{aligned}{\text{distance travelled}}&=s\left(4\right)-s\left(1\right)\\&=68-5\\&=63\\\end{aligned}[/tex]

Step 2:  

The given interval is [tex]\left[{1,4}\right][/tex] .

Now we need to find the time as,

[tex]\begin{aligned}{\text{time elapsed}}&=4-1\\&=3\\\end{aligned}[/tex]

Step 3:  

Now we find the average velocity of the particle.

The average velocity can be calculated as,

[tex]\begin{aligned}{\text{average velocity}}&=\frac{{{\text{distance travelled}}}}{{{\text{time taken}}}}\\&=\frac{{63}}{3}\\&=21{\text{units}}\\\end{aligned}[/tex]

Therefore, the average velocity of the particle is [tex]21{\text{ units}}[/tex] .

Learn more:  

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Speed, distance and time

Keywords: velocity, initial position, particle, moves, interval, distance travelled, time elapsed, position, average velocity, units, vector quantity, speed, direction, hours.

Answer:

200 texts = same costs

Step-by-step explanation:

Plan A:   20 + (.05 x 200)

              20 +  10 =30

Plan B  10 + (.10 x 200)

            10 + 20 = 30