What is the surface area of the triangular prism? *

Mathematics

1 answer:

MissTica2 years ago

5 0

The correct answer to this question is 6 in

You might be interested in

If A's income is 25% less than that of B, then how much percent is B's income more than that of A?

saul85 [17]

Let's say that B makes $100.

Then A makes $75.

So, your question then becomes 100 is what percent of 75?

This can be solved by setting up the proportion 100 / 75 = x / 100

75x = 10000

x = 133.3

So, B's income is 33.3% more than A's income.

7 0

3 years ago

Eva has £6.05 in her moneybox.

PIT_PIT [208]

Answer:

11 of 20p, 22 of 10p and 33 of 5p

Step-by-step explanation:

Eva has 20p, 10p and 5p coins, total of £6.05 = 605p

Let 20p=x, 10p=y, 5p=z

Then:

20x + 10y + 5z = 605
y : x = 2 : 1 ⇒ x= y/2
y : z = 2 : 3 ⇒ z= 3y/2

Rewriting the first equation considering next two:

10y + 10y + 7.5y = 605
27.5y = 605
y= 605/27.5
y= 22
x= y/2 = 22/2 = 11
z = 3y/2 = 3*11 = 33

Answer: 11 of 20p coins, 22 of 10p coins and 33 of 5p coins

3 0

2 years ago

Show all work and reasoning

Natalija [7]

Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

The length of the interval is $5-2=3$ , so the length of each subinterval would be $\dfrac3n$ . This means the first rectangle's height would be taken to be $x^2$ when $x=2+\dfrac3n$ , so that the height is $\left(2+\dfrac3n\right)^2$ , and its base would have length $\dfrac{3k}n$ . So the area under $x^2$ over the first subinterval is $\left(2+\dfrac3n\right)^2\dfrac3n$ .

Continuing in this fashion, the area under $x^2$ over the $k$ th subinterval is approximated by $\left(2+\dfrac{3k}n\right)^2\dfrac{3k}n$ , and so the Riemann approximation to the definite integral is

$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

and its value is given exactly by taking $n\to\infty$ . So the answer is D (and the value of the integral is exactly 39).