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EastWind [94]
3 years ago
16

Heidi solved the equation 3(x + 4) + 2 = 2 + 5(x – 4). Her steps are below: 3x + 12 + 2 = 2 + 5x – 20 3x + 14 = 5x – 18 14 = 2x

– 18 32 = 2x 16 = x Use the drops-downs to justify how Heidi arrived at each step. Step 1: Step 2: Step 3: Step 4: Step 5:
Mathematics
2 answers:
Bezzdna [24]3 years ago
6 0

Answer:

<u>Use the drops-downs to justify how Heidi arrived at each step. </u>

Step 1:  distributive property

Step 2:  combine like terms

Step 3:  subtraction property of equality

Step 4:  addition property of equality

Step 5: division property of equality

Press the <u>THANKS</u> button Plz

saveliy_v [14]3 years ago
3 0

step 1: distributive property.

step 2: combine like terms.

step 3: subtraction property of equality.

step 4: addition property of equality.

step 5: division property of equality.  

hope this helps.

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Use four rectangles to estimate the area between the graph of the function f(x) = V3x + 5 and the x-axis on the interval[0, 4] u
yuradex [85]

Answer:

  12.123

Step-by-step explanation:

You want the area under the curve f(x) = √(3x+5) on the interval [0, 4] estimated using the left sum and four subintervals.

<h3>Riemann sum</h3>

When the interval [0, 4] is divided into four equal parts, each has unit width. That means the area of the rectangle defined by the curve and the interval width will be equal to the value of the curve at the left end of the interval.

The area we want is the sum ...

  f(0) +f(1) +f(2) +f(3)

As the attachment shows, that sum is ...

  area ≈ 12.123 . . . square units

__

<em>Additional comment</em>

The table values in the attachment are rounded to 7 decimal places. Trailing zeros are not shown. Actual values used have 12 significant digits, as the total shows.

Such a sum is called a Riemann sum, named for a German mathematician. Four such sums are commonly used, and further refinements are possible. Those are the left sum (as here), the right sum, the midpoint sum, and a sum using a trapezoidal approximation of the rectangle area.

For left, right, and midpoint sums, n function values are required for n subintervals. When the trapezoidal approximation is used, n+1 function values are required.

7 0
1 year ago
At which root does the graph of f(x) = (x + 4)(x + 7,5) cross the x-axis?<br> -7<br> -4<br> 4<br> 7
Alona [7]

Answer:

Answer: -7

Step-by-step explanation:

5 0
2 years ago
How do you solve his with working
AlexFokin [52]
Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}&#10;\textit{circumference of a circle}\\\\ &#10;2\pi r&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{arc's length}\\\\&#10;s=\cfrac{\theta r\pi }{180}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+&#10;\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}&#10;\\\\\\&#10;15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}&#10;\textit{area of a circle}\\\\ &#10;\pi r^2&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{area of a sector of a circle}\\\\&#10;s=\cfrac{\theta r^2\pi }{360}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}&#10;\\\\\\&#10;90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

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Which data set has the widest spread?
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