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Yuri [45]
2 years ago
6

What is the volume of this rectangular prism? 3 units 1 cubic unit 3 units

Mathematics
1 answer:
Scorpion4ik [409]2 years ago
7 0

Answer:

Step-by-step explanation:

3x3x3 is 46 than 46 _ 4 is 57

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Solve (6-5x)/10-(2x+1)/5+x/3=2x/15
Ulleksa [173]
Multiply2 * x/15 to 2x/15

multiply both sides by 30 which is the LCM of 10, 5, 3, and 15

expand it

simplify 18 - 15x - 12x - 6 + 10x to 12 - 17x

add 17x to both sides

add 4x + 17x to = 21x

divide both sides by 21

simplify 12/21 to 4/7

now simplify

Answer: x = 4/7.
3 0
3 years ago
A blimp, suspended in the air at a height of 600 feet, lies directly over a line from a sports stadium to a planetarium. If an a
Molodets [167]

Step-by-step explanation:

so, this sounds like the blimp is located between the stadium and the planetarium.

we have a triangle :

the ground distance between the stadium and the planetarium is the baseline.

and the 2 lines of sight from the blimp on one side to the stadium and on the other side to the planetarium are the 2 legs.

we know the height of this triangle is 600 ft.

the angle of depression down to the stadium is 37°. which makes the inner triangle angle at the ground point at the stadium also 37°.

and the angle of depression down to the planetarium is 29°. which makes the inner triangle angle at the ground point at the planetarium also 29°.

and because the sum of all angles in a triangle is always 180°, we know the angle at the blimp is

180 - 37 - 29 = 114°

in order to solve this triangle, we need to split it into 2 right-angled triangles by using the height of the main triangle as delimiter.

we get a stadium side and a planetarium side triangle.

the baselines (Hypotenuses) of the 2 triangles are the corresponding lines of sight from the blimp.

the height of the large triangle is also a height and a leg in each small triangle.

and the stadium side part of the large baseline (between ground point and intersection with the height) is the second leg for the stadium side triangle.

and correspondingly, the planetarium side part of the large baseline (between ground point and intersection with the height) is the second leg for the planetarium side triangle.

the inner blimp angle of the stadium side triangle is

180 - 37 - 90 = 53°

and the inner blimp angle of the planetarium side triangle is

180 - 29 - 90 = 61°

now we can use the law of sine to get the lengths of the 2 parts of the baseline of the large triangle.

and when we add these 2 numbers we get the distance between stadium and planetarium.

law of sine is

a/sin(A) = b/sin(B) = c/sin(C)

with the sides being opposite of the associated angles.

for the stadium side triangle we get

part1/sin(53) = 600/sin(37)

part1 = 600×sin(53)/sin(37) = 796.226893... ft

for the planetarium side triangle we get

part2/sin(61) = 600/sin(29)

part2 = 600×sin(61)/sin(29) = 1,082.428653... ft

the distance between the stadium and the planetarium is

part1 + part2 = 1,878.655546... ft

4 0
1 year ago
N² + 14 = 12n<br>how do I solve by quadratic formula ​
inysia [295]

Answer:

n = 6 + \sqrt{22}  or  n = 6 - \sqrt{22}

Step-by-step explanation:

We can solve this equation using the quadratic formula OR Completing the Square method.

n² + 14 = 12n

rearrange :  n² - 12n + 14  = 0  

here  a= 1 , b = -12,  c = 14

the quadratic formula says:   x =  - b/ (2a)  +  root(b^2 - 4ac) / (2a)

or  x =  - b/ (2a)  -  root(b^2 - 4ac) / (2a)

x =  - (-12)/ (2)  +  root((-12)^2 - 4*14) / (2)

x = 6  +  root (144 - 56) / 2

x = 6 + root(88)/2

x = 6 + root(4*22) / 2

x = 6 + 2*root(22)/2

x = 6 + root(22)  = 6 + \sqrt{22}

so   x =6 + \sqrt{22}   or  x = 6 - \sqrt{22}

In this case  x = n

n = 6 + \sqrt{22}  or  n = 6 - \sqrt{22}

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3 years ago
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6 more playersssssss
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Y = 48 becuase you travelled the amount of minuets 45 x
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