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r-ruslan [8.4K]
3 years ago
7

Find the volume of the composite figure. please!

Mathematics
2 answers:
Alecsey [184]3 years ago
8 0
3times2times1equals 6
7times6times1equals 42
Irina-Kira [14]3 years ago
3 0
Volume= Length * width ( aka Base ) * height 

Break up the figure into easier figures, a small square and a bigger square.

Small square- 3*2*1 = 6 cm

Large Square- 7*6*1 = 42 cm

42 + 6 = 48 cm. But wait! You have to take one more step, which is minus-ing 6 from 48. Why? Notice that there is a little area where a side of the small square meets the bigger square. That little area is worth 3 cm ( length is 3, height is 1 cm ) times 2 ( 3 cm is one side, another 3 cm is the other side from the other square ) = 6 cm.

Your total answer should be 42. ( or 48, if your teacher doesn't count the area where the squares meet/join together ).

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Answer:

61%

Step-by-step explanation:

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This is 61% approx.

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Move all the terms involving the sine to one side, and all the numbers to the other:

9\sin(c)-2=\sin(c)-7 \iff 8\sin(c) = -5\iff \sin(c)=-\dfrac{5}{8} \iff c = \arcsin\left(-\dfrac{5}{8}\right)\approx 321.3

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What is the mathematic problem 1 3/4 x 2
Vikentia [17]

Answer:

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Step-by-step explanation:

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The teacher passed out math books. The length is 10 inches and the width is 8 inches. What is the perimeter of each book?
loris [4]
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Consider functions of the form f(x)=a^x for various values of a. In particular, choose a sequence of values of a that converges
sleet_krkn [62]

Answer:

A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.

Step-by-step explanation:

A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):

f(x)=a^{x}\\ f'(x)=a^xln(a)

The process will help us to understand what is happening, at first we rewrite the function:

f(x)=a^x\\ f(x)=e^{ln(a^x)}\\ f(x)=e^{xln(a)}\\

And then, we use the chain rule to differentiate:

f'(x)=e^{xln(a)}ln(a)\\ f'(x)=a^xln(a)

Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know  that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).

In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):

if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b)  (and below we explain why this happen)

Considering that ln(a) is a growing function and ln(e)=1, we have:

if a<e<b, then ln(a)< 1 <ln(b)

if a<e, then aˣln(a)<aˣ

if e<b, then bˣ<bˣln(b)

And because eˣ is defined to be the same as its derivative, the cases above results in the following

if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)

if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )

if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )

but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)

3 0
3 years ago
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