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3 years ago

Someone please, HELP!!!!

Mathematics

1 answer:

Brrunno [24]3 years ago

8 0

Range: maximum - minimum
IQR: third quartile - first quartile
median: middle number once all numbers are put in numerical order

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If y varies directly as x^2 and y=12 when x=2, find y when x=5

Shkiper50 [21]

$y=ax^2\\\\y=12\ \ \ and\ \ \ x=2\ \ \ \Rightarrow\ \ \ 12=a\cdot 2^2\ \ \ \Rightarrow\ \ \ a= \frac{4}{12}= \frac{1}{3} \\\\y=\frac{1}{3} x^2\\\\x=5\ \ \ \Rightarrow\ \ \ y=\frac{1}{3} \cdot5^2=\frac{25}{3} =8\frac{1}{3}$

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3 years ago

Answer for Answer? I need the answer to 8 and 9. I’ll answer 2 of yours

steposvetlana [31]

Let's go:

8)

Total of senior students = 38
Total of senior students that preferred evening classes: 7
7/38 = aproximately 0,18 or 18%

Alternative A)

9)
Imagine being sold 100 desserts. Of these 100 desserts, 75% até chocolate, meaning 75 of these desserts are chocolate. Of these 75, 23% contains nuts. So:

75 × 0,23 = 17,25 = aproximately 17,3%

Alternative C)

I hope I helped you.

6 0

3 years ago

Read 2 more answers

sume of angles of a triangel greater than 180 degrees when teh triange is so large that it extends to cosmological scales

Vesna [10]

In hyperbolic geometry, the angle sum of a triangle is always less than 180 degrees.

A lune is a wedge of a sphere with angle θ, represented by L(θ) in the proof.

α, β, and γ are the three angles of the triangle.

4πr²+4area[αβγ]=2L(α)+2L(β)+2L(γ)

2(2πr²+2area[αβγ])=2(L(α)+L(β)+L(γ))

2πr²+2area[αβγ]=L(α)+L(β)+L(γ)

At this point, we need to use a theorem that states that a lune whose corner angle is θ radians has an area of 2θr².

2πr²+2area[αβγ]=2αr²+2βr²+2γr²

2πr²+2area[αβγ]=2r²(α+β+γ)

π+area[αβγ]r²=α+β+γ

At this point, it is clear that the sum of the angles is equal to π plus the area[αβγ]r² (which cannot be zero).

To learn more about triangles and geometry,

brainly.com/question/2773823

#SPJ4

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11 months ago

QUICK HELP!!! Daniel guesses that he scored a 95 on his science test. His actual score is 91. What is the percent error in Danie

bija089 [108]

The answer is %0.8635 hope this helps

8 0

3 years ago

Read 2 more answers

Write an equation of the perpendicular bisector of the segment with endpoints G 9,8       and H 3,2      .

Norma-Jean [14]

Answer:

y = -x + 11

Step-by-step explanation:

The equation of a straight line is is given by:

y = mx + b; where m is the slope and b is the y intercept

The equation of the line joining G(9, 8) and H(3, 2) is given as:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-8=\frac{2-8}{3-9}(x-9)\\\\y-8=x-9\\\\y=x-1$

The perpendicular bisector of the line joining G(9, 8) and H(3, 2) is perpendicular to the line joining G(9, 8) and H(3, 2) and passes through the midpoint of line joining G(9, 8) and H(3, 2).

Let (x, y) be the midpoint of the line joining G(9, 8) and H(3, 2). Hence:

x = (9 + 3)/2 = 6

y = (8 + 2)/2 = 5

The midpoint = (6, 5)

Two lines are perpendicular if the product of their slopes is -1.

The line joining G(9, 8) and H(3, 2) has a slope of 1, hence, the slope of the perpendicular bisector would be -1.

This means that the perpendicular bisector has a slope of -1 and passes through (6, 5). Using:

$y-y_1=m(x-x_1)\\\\y-5=-1(x-6)\\\\y-5=-x+6\\\\y=-x+11$

The equation of the perpendicular bisector is y = -x + 11

8 0

3 years ago