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3 years ago

Write a fraction that is less than 5/6 and has a denominator of eight

Mathematics

1 answer:

jonny [76]3 years ago

6 0

Answer: (1/8) (2/8) (3/8) (4/8) (5/8) any of these would work

Step-by-step explanation:

all of these are lower in value in comparison to 5/6

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If ,. The volume of tetrahedron whose three adjacent sides are represented by vectors is, where means angle between.

marissa [1.9K]

Answer:Given | a⃗ | = | c⃗ | = | c⃗ | = 2 Angle between a⃗ and b⃗ is pi3 a⃗·b⃗ = | a⃗ | | b⃗ |co

Step-by-step explanation:

6 0

3 years ago

How’s do I simplify this

Sati [7]

1/8x(16m-8)-17

Factor out 8 from the expression

1/8x8(2m-1)-17

Reduce the numbers with greatest common divisor 8

(2m-1)-17

Remove unnecessary parenthesis

2m-1-17

Calculate the difference

2m-18

3 0

4 years ago

Read 2 more answers

A taxi service charges an initial $15 fee, plus $3 per mile driven. A. Write an equation to represent the situation. B. How much

Paraphin [41]

A. Let x represent the number of miles and y represent the total price.

y=3x+15

B. Using the equation we just wrote, replace the x with 45.

y=3(45)+15

Now multiply 45 by 3.

y=135+15

And finally add together 135 and 15.

y=150

The total cost of a 45 mile taxi ride would be $150.00.

4 0

3 years ago

Read 2 more answers

The histogram shows informstion about how 550 peopel travel to work

irina [24]

Answer:

1) 100 ) 15

Step-by-step explanation:

1) 30 to 50 is 20 miles so 5 * 20 is 100

2)20 to 50 is 30 miles so its 150 and 15 to 20 is 5 so 11 *5 is 55

55+150 equals 250 so the answer is 15 miles.

4 0

3 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

3 years ago