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2 years ago

Express the following quentitis as ratio 10k to 50k

Mathematics

1 answer:

alexandr402 [8]2 years ago

3 0

Answer:

watch me whipzdudbbd iv ex

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Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter

Levart [38]

Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

..............8/sqrt(64+x^2)

A. ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and $\sqrt{8^{2} - x^{2} }$ or $\sqrt{64 - x^{2} }$

thus tan(arcsin(x/8)) = $\frac{x}{\sqrt{64 - x^{2} }}$ =

Therefore ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

Here we have cos = adjacent/hypotenuse where adjacent side is $\sqrt{64 - x^{2} }$ and hypothenuse = 8 we have $\sqrt{64 - x^{2} }$ /8

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

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2 years ago

What is 3 and 1/5 as a decimal

Musya8 [376]

Answer:

3.2

Step-by-step explanation:

1/5=0.2

3 + 0.2 = 3.2

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3 years ago

A. x = 1 b. x = 2 c. x = 0 d. x = 3

ella [17]

I think d but not sure

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3 years ago

Which rule describes the composition of transformations that maps ABC to A'B'C"?

Sophie [7]

I think it’s transfiguration :)

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3 years ago

Read 2 more answers

Use Identities to find the exact value. 24) cos(-75°)

harkovskaia [24]

Answer:

$\frac{1}{4}$ ( $\sqrt{6}$ - $\sqrt{2}$ )

Step-by-step explanation:

Using the addition formula for cosine

cos(x - y) = cosxcosy + sinxsiny

and the exact values

sin45° = cos45° = $\frac{1}{\sqrt{2} }$

cos60° = $\frac{1}{2}$ , sin60° = $\frac{\sqrt{3} }{2}$

Note that

cos(- 75)° = cos(45 - 120)°, thus

cos(45 - 120)°

= cos45° cos120° + sin45° sin120°

= cos45° ( - cos60°) + sin45° sin60°

= $\frac{1}{\sqrt{2} }$ × - $\frac{1}{2}$ + $\frac{1}{\sqrt{2} }$ × $\frac{\sqrt{3} }{2}$

= - $\frac{1}{2\sqrt{2} }$ + $\frac{\sqrt{3} }{2\sqrt{2} }$

= $\frac{\sqrt{3}-1 }{2\sqrt{2} }$ × $\frac{\sqrt{2} }{\sqrt{2} }$

= $\frac{\sqrt{2}(\sqrt{3}-1) }{4}$

= $\frac{1}{4}$ ( $\sqrt{6}-\sqrt{2}$ )

6 0

3 years ago

Express the following quentitis as ratio 10k to 50k​

Express the following quentitis as ratio 10k to 50k