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3 years ago

Select from the drop-down menu to correctly identify the property shown.

Mathematics

2 answers:

lilavasa [31]3 years ago

8 0

The answer is Commutative Property.

anyanavicka [17]3 years ago

4 0

Answer:

Option B is correct.

Commutative property is used here.

Explanation:

Commutative Property states that we can add any number in any order, we will get the same result i.e, a+b = b+a.

let a = -4.8 and b = (3.2+(-1.6))

then, using commutative property we have;

a+b=b+a then,

−4.8 + (3.2 + (−1.6)) = (3.2 + (−1.6)) + (−4.8).

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What is the solution for a + 5 and two-thirds = 9

erastova [34]

Answer:

8.5

Step-by-step explanation:

Do you mean "(a+5)2/3

(a + 5) 2/3 = 9

a + 5 = 13.5

a=8.5

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3 years ago

Which of the following is the equation of a circle with a radius of 10 cm and center at (–3, 6)?

Volgvan

The equation of the circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have:

$r=10\\(-3,\ 6)\to h=-3,\ k=6$

Substitute:

$(x-(-3))^2+(y-6)^2=10^2\\\\(x+3)^2+(y-6)^2=100$

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3 years ago

A bag contains 8 3/4 pounds of cat food . How many 3/8 pound servings can be made from the bag? please answer :(

LiRa [457]

Answer: 23 1/3

Step-by-step explanation:

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3 years ago

Find the surface area of the rectangular and triangular prisms

harina [27]

Is there a picture to this????

if your just asking how to find the area of something all you need to do is multiply all the numbers together and the total is the area.
For example if its 8cm,3cm,2cm all you need to do is multiply them together
8x3x2=48 so your area is 48cm

7 0

3 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.

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

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

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.

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

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

7 0

3 years ago