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loris [4]
3 years ago
14

Use the provided ruler to find the scale and the scale factor of the drawing. The scale measures 4 centimeters equal to 120 mete

rs. The scale factor is ?
Mathematics
1 answer:
PtichkaEL [24]3 years ago
5 0

Answer:

1 : 3000

Step-by-step explanation:

Given that:

Scale measurement as following:

4 centimeters are equal to 120 meters.

To find:

The scale factor = ?

Solution:

First of all, let us learn about the definition of scale factor.

Scale factor means 1 unit on scale is equal to what length of actual measurement.

We are given that 4 cm on scale is equal to 120 meters of actual length.

We need to find nothing but how much 1 cm of scale is equal to actual length.

4 cm of scale = 120 meters or 12000 cm of actual length

Dividing by 4 on both sides, we get:

\frac{4}{4} = 1 cm of scale = \frac{12000}{4} = 3000 cm of actual length

Therefore, the scale factor is <em>1 : 3000</em>

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How can solve this ???
poizon [28]

Answer: $11,808

<u>Step-by-step explanation:</u>

2011 - 1995 = 16

418t + 5120

= 418(16) + 5120

= 6688 + 5120

= 11,808


5 0
3 years ago
a cell phone company has a basic monthly plan of $40plus $0.45 for many minutes ised over 700. John was charged a total of $48.1
jonny [76]
Cost of month plan = $40

Cost of minutes used = $0.45 after 700 minutes

Total cost = $48.10

Let x be the total minutes used
Total cost = 40 + 0.45(x - 700)

------------------------------------------------------------------
Form the equation and solve x
------------------------------------------------------------------
40 + 0.45(x - 700) = 48.10
40 + 0.45x - 315 = 48.10
0.45x - 275 = 48.10
0.45x = 48.10 + 275
0.45x = 323.10
x = 323.10 ÷ 0.45
x = 718

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Answer: John had used 718 mins this month
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8 0
4 years ago
How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y
son4ous [18]

Solution:

Given:

x^2=6y

Part A:

The vertex of an up-down facing parabola of the form;

\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}

Rewriting the equation given;

\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\  \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\  \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\  \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\  \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}

Therefore, the focus is;

(0,\frac{3}{2})

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}

Therefore, the directrix is;

y=-\frac{3}{2}

3 0
1 year ago
Somebody help me plz
larisa86 [58]
A is the correct answer
8 0
3 years ago
Read 2 more answers
Jacob walks 1.5 miles north. He turns and walks 0.8 miles west. How far is he from his starting point?
Flura [38]

Answer:

1.7 miles

Step-by-step explanation:

Given that,

Jacob walks 1.5 miles north.

He turns and walks 0.8 miles west.

We need to find how far is he from his starting point. Let he is at a distance of x miles from the starting point. It can be calculated as follows :

x=\sqrt{1.5^2+0.8^2} \\\\x=1.7\ \text{miles}

Hence, he is 1.7 miles from his starting point.

7 0
3 years ago
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