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Oduvanchick [21]
2 years ago
6

BRAINLIEST TO CORRECT HURRY

Mathematics
1 answer:
dalvyx [7]2 years ago
8 0

5 16/25 is the greatest and -18/25 is the least if I'm correct

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10/9 or 1 1/9

I suggest using an app called Photomath for equations
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Jackson hikes 220.2 meters up a mountain. If he is 60% of the way up, how tall is the mountain?
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367 meters tall
divide 220.2 by 0.6
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-2/x^2-4 + x-1/x^2-2x
Fiesta28 [93]

Answer:

I need help with this too

Step-by-step explanation:

8 0
3 years ago
ASAP!!! PLEASE help me solve this question!!!! I really need help...
GenaCL600 [577]

Answer:

304(pi) g

Step-by-step explanation:

First we find the volume of the hollow ball. Then we find the mass using the volume and density.

Let R = exterior radius = 3 cm

Let r = interior radius = 2 cm

volume = exterior volume - interior volume

volume = (4/3)(pi)R^3 - (4/3)(pi)r^3

volume = (4/3)(pi)(R^3 - r^3)

volume = (4/3)(pi)(3^3 - 2^3) cm^3

volume = (4/3)(pi)(27 - 8) cm^3

volume = (76/3)pi cm^3

Now we use the density and the volume to find the mass.

density = mass/volume

mass = density * volume

mass = 12 g/cm^3 * (76/3)pi cm^3

mass = 304(pi) g

Answer: 304(pi) g

6 0
3 years ago
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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)

B

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

4 0
2 years ago
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