You draw a line on a paper. Using your pencil, but without touching that line again, can you make the line longer?

Two sides of a triangle have the same length. The third side measures 2 m less than twice the common length. The perimeter of th

kozerog [31]

Answer:

9m is the length of both common sides

9m-2m=7m is the length of the uncommon side

9m×2+7m=26m

5 0

3 years ago

Which fraction is not in simplest form? Explain. : 9/21 , 7/18, 3/25, 12/31

Slav-nsk [51]

9/21 because it can still be reduced to 3/7

8 0

3 years ago

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In a certain card game you draw one card off a standard deck of 52 cards. If you draw a spade you get paid $12, if you draw a re

Mariulka [41]

Step-by-step explanation:

In a standard deck of 52 cards, there are 2 red aces, 2 red Queens, and 13 spades. That leaves 35 cards for everything else.

For the game to be fair, the cost must equal the expected value. The expected value is the sum of each outcome times its probability.

C = (12) (13/52) + (20) (2/52) + (38) (2/52) + (0) (35/52)

C = 68/13

C ≈ 5.2308

4 0

2 years ago

The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha

Xelga [282]

Answer:

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

Step-by-step explanation:

Let suppose that function $f$ is continuous and integrable in the given intervals, by integral definition of average we have that:

$\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6$ (1)

$\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20$ (2)

By Fundamental Theorems of Calculus we expand both expressions:

$\frac{F(3)-F(-2)}{3-(-2)} = -6$

$F(3) - F(-2) = -30$ (1b)

$\frac{F(5)-F(3)}{5-3} = 20$

$F(5) - F(3) = 40$ (2b)

We obtain the average value of $f$ over the interval $[-2, 5]$ by algebraic handling:

$F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)$

$F(5) - F(-2) = 10$

$\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}$

$\bar f = \frac{10}{7}$

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

4 0

2 years ago

a pyramid whose altitude is 5ft weighs 800lbs. at what distance from its vertex must it be cut by a plane parallel to its base s

slavikrds [6]

A structure which has a square base and four triangular sides meeting at a point is called pyramid.

At distance of 3.97 feet from its vertex , pyramid is cut by plane so that the two solids of equal weight will be formed.

It is assumed that weight of pyramid is proportional to its volume.

So, $w = k V$ , where V is volume of original pyramid and v is volume of small pyramid and k is constant.

Let us consider that at h distance, pyramid is cut from its vertex. So a small pyramid is also formed.

Assume that base area of original pyramid is A and base of small pyramid is a.

Volume of original pyramid is, $V= \frac{1}{3} *A* 5$

So, weight of original pyramid, $W = k *(\frac{1}{3} *A* 5) = 800$

Volume of small pyramid is, $v = \frac{1}{3}* a* h$

So, weight of small pyramid, $w = k*(\frac{1}{3}* a* h)=400$

Since, base and height of small pyramid and original pyramid are in proportion.

So, $\frac{a}{A} = (\frac{h}{5}) ^{2}$

$a = (\frac{h}{5} )^{2}A$

Substituting value of a in equation $k*(\frac{1}{3}* a* h)=400$

So, $k*(\frac{1}{3}* \frac{h^{2} }{25}A * h)=400$

$(k*\frac{1}{3}* A*5)*\frac{1}{5} *\frac{h^{2} }{25} * h)=400$

Since, $(k *\frac{1}{3} *A* 5) = 800$ , substitute in above equation.

So, $800*\frac{h^{3} }{125}=400\\\\h^{3}=\frac{125}{2}\\\\h=\sqrt[\frac{1}{3} ]{62.5}\\\\h=3.97 feet$

Learn more:

brainly.com/question/17950304

6 0

2 years ago

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