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Andrew [12]
3 years ago
8

Serenity wanted to make a lot of bows that she could sell. She had 6 1/4 yards of ribbon. Each bow takes 3/4 yard to make. How m

any bows could she make? How much extra fabric was left?
Mathematics
1 answer:
Gnesinka [82]3 years ago
7 0

Answer:

8 Bows

Step-by-step explanation:

I will be using money as an example.

It takes 4 quarters to make a dollar. (1/4 1/4 1/4 1/4 = 1)

By using this method we will know that it takes 24 quarters to make 6 dollars and the additional 1/4 is an extra quarter. Now that we have 25 quarters we will divide it by 3 because it takes 3/4 to make a bow.

6x4=24

24 + 1 = 25

25/3= 8.333333

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The graph of the derivative of a function f crosses the x-axis 3 times. What does this tell you about the graph of f ?
expeople1 [14]

Answer:

D. The function f has 3 horizontal tangent lines

Step-by-step explanation:

Well whenever any function crosses the x-axis, it means that the y-value is equal to zero.

In this case when the derivative passes the x-axis it indicates the tangent line corresponding to that x-value has a slope of zero, which when plotted is a horizontal line.

This means that the graph f has 3 horizontal tangent lines.

7 0
1 year ago
Read 2 more answers
What is a simplified expression for the perimeter of the polygon?
serious [3.7K]

Answer:

D. 5x^2 + 2y^2 + 12x + 5y

Step-by-step explanation:

6x + 2y + y^2 + 6x + 2y + y^2 + 4x^2 + x^2 + y

6x + 6x = 12x

12x + 2y + y^2 + 2y  + y^2 + 4x^2 + x^2 + y

2y + 2y + y = 5y

12x + 5y + y^2 + y^2 + 4x^2 + x^2

y^2 + y^2 = 2y^2

2y^2 + 4x^2 + x^2 + 12x + 5y

4x^2 + x^2 = 5x^2

2y^2 + 5x^2  + 12x + 5y

5x^2 + 2y^2 + 12x + 5y

7 0
3 years ago
Read 2 more answers
Question 5 please and thank you
stira [4]
Need a more clear photo please can’t see nothing.
6 0
3 years ago
Which is not an equation of the line going through (3, -6) and (1, 2)?
777dan777 [17]

Answer:

B, C

Step-by-step explanation:

The equation of a line can be given by y -y₁= m(x -x₁), where m is the slope. This is also known as the point-slope form.

\boxed{ slope = \frac{y _{1} - y_2 }{x_1 - x_2} }

Slope of the line

=  \frac{2 - ( - 6)}{1 - 3}

=  \frac{2 + 6}{ - 2}

= \frac{8}{ - 2}

= -4

<em>Substitute m= -4 into the equation:</em>

y -y₁= -4(x -x₁)

<em>Substitute a pair of coordinates into (x₁, y₁):</em>

Let's start by substituting (1, 2).

y -2= -4(x -1)

This gives us the same equation as D, making D an incorrect option. Note that the question asks for which is not the correct equation.

Let's change the above into the slope-intercept form, where by y is the subject of formula.

<em>Start by expanding the right-hand side:</em>

y -2= -4x +1

<em>+</em><em>2</em><em> on both sides:</em>

y= -4x +3

This equation is not the same as C. C is thus the correct option.

Let's check for options A and B.

The equation in option B is not the correct equation either as they have substituted (2, 1) instead of (1, 2) into (x₁, y₁). Thus, option B is also correct.

y -y₁= -4(x -x₁)

Substitute (3, -6) into (x₁, y₁):

y -(-6)= -4(x -3)

y +6= -4(x -3)

This is the same as option A, making option A incorrect too.

4 0
2 years ago
If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,
julsineya [31]
<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

<h2>Why?</h2>

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)

(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0
4 years ago
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