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Sunny_sXe [5.5K]
3 years ago
10

Each base of a triangular prism is a right triangle with side lengths 3 inches, 4 inches, and 5 inches. The height of the prism

is 7 inches. What is the surface area of the prism?
Mathematics
1 answer:
Artyom0805 [142]3 years ago
6 0
Well, each base has 3*4/2=6 square inches. The lateral faces have 3*7, 4*7 and 5*7 as areas, totalling 12*7=84 square inches on the sides. The bases which total 12 sq. in. are added to the sides to get a total surface area of 96 square inches.
You might be interested in
If f(x) = 5x2 − 2x, 0 ≤ x ≤ 3, evaluate the Riemann sum with n = 6, taking the sample points to be right endpoints. R6 =
Leona [35]

Answer:

46.375

Step-by-step explanation:

Given information:

f(x)=5x^2-2x

where, 0 ≤ x ≤ 3.

We need to divde the interval [0,3] in 6 equal parts.

The length of each sub interval is

\dfrac{b-a}{n}=\dfrac{3-0}{6}=0.5

Right end points are 0.5, 1, 1.5, 2, 2.5, 3.

The value function on each right end point are

f(0.5)=5(0.5)^2-2(0.5)=0.25

f(1)=5(1)^2-2(1)=3

f(1.5)=5(1.5)^2-2(1.5)=8.25

f(2)=5(2)^2-2(2)=16

f(2.5)=5(2.5)^2-2(2.5)=26.25

f(3)=5(3)^2-2(3)=39

Riemann sum:

\sum_{n=1}^6 f(x_n)\Delta x_n

Sum=[f(0.5)+f(1)+f(1.5)+f(2)+f(2.5)+f(3)]\times 0.5

Sum=[0.25+3+8.25+16+26.25+39]\times 0.5

Sum=92.75\times 0.5

Sum=46.375

Therefore, the Riemann sum with n = 6 is 46.375.

6 0
2 years ago
Q:
Neporo4naja [7]
a_1=200;\ a_2=196;\ a_3=192\\\\r=a_2-a_1\to r=126-200=-4\\\\a_n=a_1+(n-1)r\\\\a_n=200+(n-1)\cdot(-4)=200-4n+4=204-4n\\\\a_{14}=204-4\cdot14=204-56=148\\\\Answer:A
3 0
3 years ago
this picture is a scale drawing of a basketball court and 1 inch on the picture represents 15 feet of actual length. the dementi
bekas [8.4K]
If one inch represents 15 feet, then:

2 inches represent:
2 * 15 = 30 feet

1 2/3 inches represent:
1 2/3 * 15 =
= 5/3 * 15 =
= 75/3 =
= 25 feet

So the real area of the court is 30 * 25 = 750 square feet.
5 0
3 years ago
write an equation of the line in slope-intercept form. and then there is a graph that has two points on it, (-1, 3) and (0, 4)
Georgia [21]
(y-3)/(x-(-1))=(4-3)/(0-(-1))
y=x+4
7 0
3 years ago
3. The weights of apples in a shipment from Al's Orchards are normally distributed with a mean of 142 grams and a standard devia
Kamila [148]
The z scores corresponding to the mean weight of a rotten apple relative to the apples from either orchard are, respectively,

\dfrac{142-156}9\approx-1.56

\dfrac{165-156}{13}\approx0.69

The mean for the standard normal distribution is 0, which means that z scores closer to 0 represent data points that are more likely to occur. Therefore it's reasonable to believe that rotten apples occur more often in shipments from Zippy's.
3 0
3 years ago
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