All categories

3 years ago

Did I do this correctly?

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

6 0

Answer:

No, The answer should be 240

Step-by-step explanation:

The first step to finding the Surface Area was done well and was right but you got confused when trying to add the area of each shape
There is 3 Rectangles and 2 Traingles.
Let's start with the area of the triangles. 1/2bh is the formula for finding area of Traingles. 1/2(4)(6) = 12. We have two Traingles so that would be 24.
Now we have three rectangles with different lengths we need to calculate.
Rectangle 1 = 12 * 8 = 96
Rectangle 2 = 12 * 6 = 72
Rectangle 3 = 12* 4 = 48
24+96+72+48 = 240.
So, the surface area of the rectangular prisim is 240

You might be interested in

A t-shirt increased in price by

AVprozaik [17]

Step-by-step explanation:

81×1/4

=20.24

this is the answer

5 0

3 years ago

A weightlifter holds a 1,700 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the ba

masha68 [24]

In physics, work is equal to the product of force and distance, given that the direction of force is the same with the direction of the distance.

W = Fd

The total force is
F = 1,700 N + 500 N = 2,200 N
Now, if the question is the work needed to lift the barbell from the ground, the distance would be 2 m. Hence, the work would be

W = (2,220 N)(2 m) = 4,400 N·m or 4,400 J

5 0

3 years ago

Find the nth degree polynomial function with real coefficients satisfying the given conditions.

RUDIKE [14]

$\bf (x+1)(x-3)(x-2-4i)(x-2+4i)
\\\\
\textit{now, let's take a peek at }(x-2-4i)(x-2+4i)\\
\textit{and recall our }\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
(x-2-4i)(x-2+4i)\implies [(x-2)-(4i)][(x-2)+(4i)]
\\\\\
[(x-2)^2-(4i)^2]\implies [(x^2-2x+4)-(4^2\boxed{i^2})]
\\\\\
[(x^2-2x+4)-(16\cdot \boxed{-1})]\implies [(x^2-2x+4)+16]
\\\\
(x^2-2x+20)\\\\
-----------------------------$

$\bf thus
\\\\

\begin{array}{llll}
(x+1)(x-3)\\(x-2-4i)(x-2+4i)
\end{array}\implies (x+1)(x-3)(x^2-2x+20)
\\\\\\
(x^2-2x-3)(x^2-2x+20)\implies 
\begin{array}{llll}
x^4-2x^3+20x^2-2x^3\\+4x^2-40x+3x^2-6x+60
\end{array}
\\\\\\
x^4-4x^3+27x^2-46x+60$

4 0

3 years ago

Find the equation of the line that passes through (-2,3) and (4,-1).

horsena [70]

$Genera\ equation\ for\ line\\\\y=ax+b\\\\To\ find\ a\ and\ b\ substitude\ points\ (-2,3),\ (4,-1)\ into\ equation\\\\ \left \{ {{3=-2a+b}\ \ \ \ \atop {-1=4a+b\ |*-1}} \right.\\\\ \left \{ {{3=-2a+b}\ \ \ \ \atop {1=-4a-b\}} \right.\\+----\\\\Addition\ method\\\\4=-6a\ \ |:-6\\\\a=-\frac{4}{6}=-\frac{2}{3}\\\\b=3+2a=3+2* (-\frac{2}{3})=3-\frac{4}{3}=1\frac{2}{3}\\\\y=-\frac{2}{3}x+1\frac{2}{3}$

8 0

3 years ago

Identify the reference angle ∅ for each given angle, 0.

frutty [35]

Answer:

When Ø = 300°, Ø = 60 degrees.

When Ø = 225°, Ø = 45 degrees.

When Ø = 480°, Ø = 60 degrees.

When Ø = -210°, Ø = 30 degrees.

Step-by-step explanation:

Reference angles are in Quadrant I (0° to 90°).

1. Find 300° (Quadrant IV) on the unit circle. Since it's in Quadrant IV, you use 360 - 300 = 60° to get your answer.

2. Find 225° (Quadrant III) on the unit circle. Since it's in Quadrant III, you use 225 - 180 = 45° to get your answer.

3. The angle 480° is not on the unit circle. To find its corresponding angle between 0° and 360°, use 480 - 360 = 120°. Then, find 120° (Quadrant II) on the unit circle. Since it's in Quadrant II, you use 180 - 120 = 60° to get your answer.

4. The angle -210° is not on the unit circle. To find its corresponding angle between 0° and 360°, use -210 + 360 = 150°. Then, find 150° (Quadrant II) on the unit circle. Since it's in Quadrant II, you use 180 - 150 = 30° to get your answer.

5 0

3 years ago

Did I do this correctly?​

Did I do this correctly?