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jekas [21]
3 years ago
5

If a pond is 6m by 15m, will 40m of fencing will be enough to fenced a pond

Mathematics
1 answer:
olchik [2.2K]3 years ago
8 0

6m + 6m + 15m + 15m = 42m

no 40m will not be enough

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Josh works h hours in a week. To find his weekly pay he multiplies his hours by his pay of $30 per hour and adds $5 for his sale
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3 0
3 years ago
The volume of the triangular prism is 12.5 m³. The length is 2.5 m and the base is 2 m. What is the width?
Fantom [35]

Answer:

  (4)  5 m

Step-by-step explanation:

You want the length of side x of a right triangular prism with base edge lengths of 2.5 m and 2 m, and a volume of 12.5 m³.

<h3>Volume</h3>

The volume of the prism is given by the formula ...

  V = Bh

where B is the area of the base:

  B = 1/2bh . . . . where b and h are the leg dimensions of the right triangle

Using these formulas together, we have ...

  V = 1/2(2.5 m)(2 m)x

  12.5 m³ = 2.5x m²

Dividing by 2.5 m², we find x to be ...

  (12.5 m³)/(2.5 m²) = x = 5 m

The dimension labeled x has length 5 meters.

6 0
1 year ago
The vertex of a parabola is (0,0)and the focus is (1/8, 0) . What is the equation of the parabola?
Aleonysh [2.5K]

The equation of a parabola whose vertex is (0, 0) and focus is (1 / 8, 0) is equal to x = 2 · y².

<h3>How to derive the equation of the parabola from the locations of the vertex and focus</h3>

Herein we have the case of a parabola whose axis of symmetry is parallel to the x-axis. The <em>standard</em> form of the equation of this parabola is shown below:

(x - h) = [1 / (4 · p)] · (y - k)²     (1)

Where:

  • (h, k) - Coordinates of the vertex.
  • p - Distance from the vertex to the focus.

The distance from the vertex to the focus is 1 / 8. If we know that the location of the vertex is (0, 0), then the <em>standard</em> form of the equation of the parabola is:

x = 2 · y²     (1)

The equation of a parabola whose vertex is (0, 0) and focus is (1 / 8, 0) is equal to x = 2 · y².

To learn more on parabolae: brainly.com/question/4074088

#SPJ1

5 0
2 years ago
Find the six trig function values of the angle 240*Show all work, do not use calculator
-BARSIC- [3]

Solution:

Given:

240^0

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

sin240^0=sin(180+60)

Using the trigonometric identity;

sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny

Hence,

\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

cos240^0=cos(180+60)

Using the trigonometric identity;

cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny

Hence,

\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\  \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

tan240^0=tan(180+60)

Using the trigonometric identity;

tan(180+x)=tan\text{ }x

Hence,

\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\  \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}

To get cosec 240 degrees:

\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\  \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\  \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}

To get sec 240 degrees:

\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\  \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\  \\ Thus, \\ sec240^0=-2 \end{gathered}

To get cot 240 degrees:

\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\  \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\  \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\  \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}

5 0
1 year ago
I NEED HELP!!! 12 PTS!!!
Agata [3.3K]
The answer is 13 packages.
6 0
3 years ago
Read 2 more answers
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