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Zina [86]
3 years ago
10

The scale for a scale drawing is 8 millimeters: 1 centimeter. Which is larger,the actual object or the scale drawing? PLEASE PLE

ASE PLEASE EXPLAIN!!!
Mathematics
1 answer:
Daniel [21]3 years ago
5 0
The centimeter is bigger
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Find the equation of the axis of symmetry of the following parabola algebraically.
mr Goodwill [35]

Answer:

the equation of the axis of symmetry is x=8

Step-by-step explanation:

Recall that the equation of the axis of symmetry for a parabola with vertical branches like this one, is an equation of a vertical line that passes through the very vertex of the parabola and divides it into its two symmetric branches. Such vertical line would have therefore an expression of the form: x=constant, being that constant the very x-coordinate of the vertex.

So we use for that the fact that the x position of  the vertex of a parabola of the general form: y=ax^2+bx+c, is given by:

x_{vertex}=\frac{-b}{2\,a}

which in our case becomes:

x_{vertex}=\frac{-b}{2\,a} \\x_{vertex}=\frac{48}{2\,(3)} \\x_{vertex}=\frac{48}{6} \\x_{vertex}=8

Then, the equation of the axis of symmetry for this parabola is:

x=8

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3 years ago
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77julia77 [94]
H would be your answer (9x2)+(14x4)=74
8 0
3 years ago
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What is 100 and a long division divided by six
kolbaska11 [484]

Answer: about 16

Step-by-step explanation:

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6 0
3 years ago
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Find the values of the missing angles<br><br> HELP PLEASE DO ANY! I’ll give you brilliance
yan [13]

Answer:

1) x = 42 ; y = 42

2) x = 72 ; y = 48 ; z = 60

Step-by-step explanation:

1)

x + 42 + 96 = 180

x + 138 = 180

x = 42

y + 42 + 96 = 180

y + 138 = 180

y = 42

2)

x + 108 = 180

x = 72

y + 132 = 180

y = 48

z + x + y = 180

z + 72 + 48 = 180

z + 120 = 180

z = 60

3 0
3 years ago
Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y
ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; (x^2-4)^2y'' + (x + 2)y' + 7y = 0

The above equation can be better expressed as:

y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

p(x) = \dfrac{(x+2)}{(x^2-4)^2} \

p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \

p(x) = \dfrac{1}{(x+2)(x-2)^2}

Also;

q(x) = \dfrac{7}{(x^2-4)^2}

q(x) = \dfrac{7}{(x+2)^2(x-2)^2}

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

\lim \limits_{x \to-2} (x+ 2) p(x) =  \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{1}{(x-2)^2}

\implies \dfrac{1}{16}

\lim \limits_{x \to-2} (x+ 2)^2 q(x) =  \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{7}{(x-2)^2}

\implies \dfrac{7}{16}

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0
3 years ago
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