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3 years ago

EMERGENCY! How many cubes with side lengths of 1/4 cm does it take to fill the prism?

Mathematics

1 answer:

Fantom [35]3 years ago

3 0

Answer:

135 cubes because if you multiply all the sides you get 2 7/64 so then you multiply that with 1/4 and you get 135/256. Just use the top number. I hope this helps!

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What is the percent of 16/25

Lelechka [254]

16/25 = 0.64 = 64%

Another way to do it is to multiply top and bottom by 4
16/25 = (16*4)/(25*4) = 64/100 = 64%

Either way, the answer is 64%

7 0

3 years ago

Read 2 more answers

I need help with no.7

ziro4ka [17]

Answer:

(a) 315°

(b) 3°

Step-by-step explanation:

Bearings are measured clockwise from north. The triangle described is illustrated in the attachment.

The bearing of P from R is 180° different from the bearing of R from P it will be ...

135° +180° = 315° . . . . bearing of P from R

The bearing of Q from R is 48° more than the bearing of P from R, so is ...

315° +48° = 363°, or 3° . . . . bearing of Q from R

The angle QPR has a value that makes the sum of angles in the triangle equal to 180°. It is ...

180° -48° -55° = 77°

The bearing of Q from P is 77° less than the bearing of R from P, so is ...

135° -77° = 58°

As above, the reverse bearing from Q to P is ...

58° +180° = 238° . . . . bearing of P from Q

8 0

3 years ago

Find the derivative of f(x) = (1 + 6x2)(x − x2) in two ways.

ipn [44]

Using product rule;

f(x)=(1+6x²)(x-x²)

f'(x)=(12x)(x-x²) + (1-2x)(1+6x²) = 12x² -12x³ +1 +6x² -2x -12x³ = -24x³ +18x² -2x +1

Solving the bracket first;

f(x)=(1+6x²)(x-x²) = x -x² +6x³ -6x^4

f'(x)= 1 -2x +18x² -24x³ = -24x³ +18x² -2x +1

3 0

3 years ago

If f(x)=x^3-x+2, then (f^-1)'(2)

yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

$f\left(f^{-1}(x)\right) = x$

$\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x$

which is a cubic polynomial in $f^{-1}(x)$ with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0 ⇒ x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

$\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}$

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0 or x = 1 or x = -1

Then $\left(f^{-1}\right)'(2)$ can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0

2 years ago

John works two jobs. He earns $10 an hour babysitting his neighbor's children and $15 an hour mowing lawns in his neighborhood.

Jlenok [28]

Add up all the numbers

7 0

3 years ago