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Anon25 [30]
3 years ago
11

What does 5/6 equal as a decimal?

Mathematics
2 answers:
AlexFokin [52]3 years ago
6 0
Hi there! To turn a fraction into a decimal we divide the numerator by the denominator. So, we have the fraction 5/6 where 5 is the numerator and 6 is the denominator. 5÷6=0.8333 repeating. Therefore, 5/6 as a decimal is 0.8333
ch4aika [34]3 years ago
6 0
The decimal of 5/6 is 0.8333
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Determine whether or not 630 is a triangular number. start with the formula t n = n ( n + 1 ) 2 and then use the quadratic formu
ziro4ka [17]
Triangular sequence = n(n + 1)/2

If 630 is a triangular number, then:

n(n + 1)/2  = 630

Then n should be a positive whole number if 630 is a triangular number.

n(n + 1)/2  = 630

n(n + 1)  = 2*630

n(n + 1)  = 1260

n² + n = 1260

n² + n - 1260 = 0

By trial an error note that 1260 = 35 * 36

n² + n - 1260 = 0

Replace n with 36n - 35n

n² + 36n - 35n - 1260 = 0

n(n + 36) - 35(n + 36) = 0

(n + 36)(n - 35) = 0

n + 36 = 0   or   n - 35 = 0

n = 0 - 36,   or  n = 0 + 35

n = -36, or 35

n can not be negative. 

n = 35 is valid.

Since n is a positive whole number, that means 630 is a triangular number.

So the answer is True.
7 0
3 years ago
Find the vertex of the graph of the function.
lapo4ka [179]

Answer: The correct options are  (1) (5,10), (2) (3,-3), (3) x = -1, (4) y=(x+2)^2+3, (5) 21s and (6) 0, -1, and 5.

Explanation:

Te standard form of the parabola is,

f(x)=a(x-h)^2+k        .....(1)

Where,  (h,k) is the vertex of the parabola.

(1)

The given equation is,

f(x)=(x-5)^2+10

Comparing this equation with equation (1),we get,

h=5 and k=10

Therefore, the vertex of the graph is (5,10) and the fourth option is correct.

(2)

The given equation is,

f(x)=3x^2-18x+24

f(x)=3(x^2-6x)+24

To make perfect square add (\frac{b}{2a})^2, i.e., 9. Since there is factor 3 outside the parentheses, so subtract three times of 9.

f(x)=3(x^2-6x+9)+24-3\times 9

f(x)=3(x-3)^2-3

Comparing this equation with equation (1),we get,

h=3 and k=-3

Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.

(3)

The given equation is

f(x)=4x^2+8x+7

f(x)=4(x^2+2x)+7

To make perfect square add (\frac{b}{2a})^2, i.e., 1. Since there is factor 4 outside the parentheses, so subtract three times of 1.

f(x)=4(x^2+2x+1)+7-4

f(x)=4(x+1)^2+3

Comparing this equation with equation (1),we get,

h=-1 and k=3

The vertex of the equation is (-1,3) so the axis is x=-1 and the correct option is 2.

(4)

The given equation is,

y=x^2+4x+7

To make perfect square add (\frac{b}{2a})^2, i.e., 2^2.

f(x)=x^2+4x+4+7-4

f(x)=x^2+4x+4+7-4

f(x)=(x+2)^2+3

Therefore, the correct option is  4.

(5)

The given equation is,

h=-16t^2+672t

It can be written as,

h=-16(t^2-42t)

It is a downward parabola. so the maximum height of the function is on its vertex.

The x coordinate of the vertex is,

x=\frac{b}{2a}

x=\frac{42}{2}

x=21

Therefore,  after 21 seconds the projectile reach its maximum height and the correct option is first.

(6)

The given equation is,

f(x)=3x^3-12x^2-15x

f(x)=3x(x^2-4x-5)

Use factoring method to find the factors of the equation.

f(x)=3x(x^2-5x+x-5)

f(x)=3x(x(x-5)+1(x-5))

f(x)=3x(x-5)(x+1)

Equate each factor equal to 0.

x=0,-1,5

Therefore, the zeros of the given equation is 0, -1, 5 and the correct option is 2.

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3 years ago
Name the edges of an hexagonal pyramid​
aleksandrvk [35]

Given:

Hexagonal pyramid

To find:

The edges of an hexagonal pyramid.

Solution:

Edges means lines which connecting to vertices.

Edges in the base of the pyramid:

AB, BC, CD, DE, EF, FA

Edges in the triangular shape of the pyramid:

AG, BG, CG, DG, EG, FG

Therefore edges of an hexagonal pyramid are:

AB, BC, CD, DE, EF, FA, AG, BG, CG, DG, EG, FG

4 0
3 years ago
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frozen [14]
1. NIL
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6 0
2 years ago
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weqwewe [10]

Answer:

40.82 is the surface area of this solid.

8 0
3 years ago
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