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ElenaW [278]
3 years ago
11

Which is true about the volume or surface area of these prisms?

Mathematics
1 answer:
pshichka [43]3 years ago
5 0
Your answer is the second option: The surface area of B is greater than the surface area of A.

To find the surface area, I found the area of the three sides shown and added them together, and then multiplied the answer by 2, so:

B:
8 × 2 = 16
8 × 10 = 80
10 × 2 = 20
+
= 116
116 × 2 = 232 cm²

A:
5 × 8 = 40
5 × 4 = 20
8 × 4 = 32
+
= 92
92 × 2 = 184 cm²

So therefore B > A because 232 > 184

I hope this helps!
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At Bridgeview School,7/10 of the total students are driven to school by their parents. Another 1/6 of the total students at Brid
jarptica [38.1K]

Answer: 2/15

Step-by-step explanation:

We are given the information that at Bridgeview School, 7/10 of the total students are driven to school by their parents and that 1/6 of the total students at Bridgeview School ride the bus.

The fraction of the students that walk to school would be:

= 1 - (7/10 + 1/6)

= 1 - (21/30 + 5/30)

= 1 - 26/30

= 4/30

= 2/15

5 0
3 years ago
X^2+X=56<br> Help please!!!!
stich3 [128]
The value of x is either +7 or -8
3 0
3 years ago
18x + 24 = 12x +14. Find the value of (x)​
densk [106]

Answer:

X = -10/6

X = -1,666666666666667

Step-by-step explanation:

First at all you join the X's

18x-12x = 14-24

Then you simplified:

6x = -10

X = -10/6

X = -1,666666666666667

3 0
2 years ago
Read 2 more answers
Pls, help me with this math question
olya-2409 [2.1K]

Answer:

Perimeter \sqrt{26}+\sqrt{5}+\sqrt{10}+\sqrt{17} units.  Area 12 square units.

Step-by-step explanation:

Perimeter: total distance around the figure.

Distance Formula:  the distance between points \left(x_1,y_1\right) \text{ and } \left(x_2,y_2\right) is

d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

AB=\sqrt{(6-1)^2+(2-1)^2}=\sqrt{25+1}=\sqrt{26}

BC=\sqrt{(5-6)^2+(4-2)^2}=\sqrt{1+4}=\sqrt{5}CD=\sqrt{(2-5)^2+(5-4)^2}=\sqrt{9+1}=\sqrt{10}DA=\sqrt{(1-2)^2+(1-5)^2}=\sqrt{1+16}=\sqrt{17}

The perimeter is the sum of all those segment lengths.

One way to find the area of the figure is to surround it with a rectangle, insert some lines so that the areas you do not want can be found and subtracted from the rectangle's area.  (See attached image.)

The area of the large rectangle around the figure is 5 x 4 = 20 square units.

The triangles have areas 1/2 (base) (height):

A. (1/2)(1)(4) = 2 square units

B. (1/2)(3)(1) = 1.5 square units

D. (1/2)(1)(2) = 1 square unit

E. (1/2)(5)(1) = 2.5 square units

Square C.  (1)(1) = 1 square unit

Total of all the area you don't want to include:

2 + 1.5 + 1 + 2.5 + 1 = 8 square units

Subtract 8 from the surrounding rectangle's area of 20, and you get the area of the figure is 20 - 8 = 12 square units.

4 0
3 years ago
(-1 + 6) Squared + (4+5) squared
nalin [4]
(-1 + 6)^2 <span>+ (4+5)^2 = </span>106
3 0
3 years ago
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