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Len [333]
3 years ago
13

You are planning to plant a triangular garden in your backyard, as shown. You plan to put up a fence around the garden to keep o

ut animals. Find the length of fencing you need to the nearest meter.​

Mathematics
1 answer:
Drupady [299]3 years ago
6 0

Answer:

24 m

Step-by-step explanation:

The lenght of fencing needed is the sum of all sides the triangular garden represented on the coordinate grid.

The lenght of the triangular garden = 16 - 10 = 6 m

Lenght of the height = 8 - 0 = 8 m

Lenght of the hypotenuse can be calculated using coordinates of the two vertices of the ∆ that forms the hypotenuse lenght and also using the distance formula.

Coordinates of the two vertices = (10, 0) and (16, 8).

distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

(10, 0) = (x_1, y_1)

(16, 8) = (x_2, y_2)

d = \sqrt{(16 - 10)^2 + (8 - 0)^2}

d = \sqrt{(6)^2 + (8)^2}

d = \sqrt{36 + 64} = \sqrt{100}

d = 10

Length of fencing = 6 + 8 + 10 = 24 m

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Answer: it will be 45, but for some reason it’s not in your answer choices

Step-by-step explanation:

3 0
3 years ago
Lucy is raising money for a school trip by selling lollipops and bags of
sukhopar [10]
<h3>16 lollipops and 9 bags of chips are sold</h3>

<em><u>Solution:</u></em>

Let "x" be the number of lollipops sold

Let "y" be the number of bags of chips sold

Cost of 1 lollipop = $ 1.50

Cost of 1 bag of chip = $ 1.75

<em><u>Yesterday Lucy made $39.75 from selling a total of 25  lollipops and bags of chips</u></em>

Therefore,

x + y = 25

y = 25 - x -------- eqn 1

<em><u>Yesterday Lucy made $39.75</u></em>

Therefore,

number of lollipops sold x Cost of 1 lollipop + number of bags of chips sold x Cost of 1 bag of chip = 39.75

x \times 1.50 + y \times 1.75 = 39.75

1.5x + 1.75y = 39.75 -------- eqn 2

<em><u>Substitute eqn 1 in eqn 2</u></em>

1.5x + 1.75(25 - x) = 39.75

1.5x + 43.75 - 1.75x = 39.75

0.25x = 4

<h3>x = 16</h3>

<em><u>Substitute x = 16 in eqn 1</u></em>

y = 25 - 16

<h3>y = 9</h3>

Thus 16 lollipops and 9 bags of chips are sold

7 0
3 years ago
HELP ASAP!!ANSWER USING GIVEN FRACTIONS!!FIRST CORRECT ANSWER WILL GET BRAINLIEST!!
Leto [7]

Answer:

1/15,2/45, 4/45, 1/45

Step-by-step explanation:

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7 0
3 years ago
As a salesperson you get $18/day and $2.80. for each sale you make Ifyou want to make at least $79 in one day, how many sales wo
jonny [76]
B.22 If you take $79 subtract the $18 you will already get, you get 61 divide by 2.80 you’ll get 21.7 round up and you need to make at least 22 sales to get $79 dollars that day.
6 0
4 years ago
Divide 9x ^ 3 + 18x ^ 2 - 13x + 5 by 3x - 1 using division and write the division in the form P = DQ + R
lana66690 [7]

Answer:

\frac{9x^3\:+\:18x\:^2\:-\:13x\:+\:5}{3x-1}=9x^3+18x^2-13x+5

Step-by-step explanation:

DIVISION ALGORITHM: If p(x) and d(x)\neq 0  are polynomials, and the degree of d(x) is less than or equal to the degree of f(x),  then there exist unique polynomials q(x) and r(x), so that

                                               \frac{p(x)}{d(x)} =q(x)+\frac{r(x)}{d(x)}

and so that the degree of r(x)  is less than the degree of d(x).

To find \frac{9 x^{3} + 18 x^{2} - 13 x + 5}{3 x - 1} you must:

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}9x^3+18x^2-13x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{9x^3}{3x}=3x^2

\mathrm{Quotient}=3x^2

\mathrm{Multiply\:}3x-1\mathrm{\:by\:}3x^2:\:9x^3-3x^2\\\mathrm{Subtract\:}9x^3-3x^2\mathrm{\:from\:}9x^3+18x^2-13x+5\mathrm{\:to\:get\:new\:remainder}\\

\mathrm{Remainder}=21x^2-13x+5

Therefore,

\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+\frac{21x^2-13x+5}{3x-1}

\mathrm{Divide}\:\frac{21x^2-13x+5}{3x-1}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}21x^2-13x+54\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{21x^2}{3x}=7x\\\\\mathrm{Quotient}=7x

\mathrm{Multiply\:}3x-1\mathrm{\:by\:}7x:\:21x^2-7x\\\mathrm{Subtract\:}21x^2-7x\mathrm{\:from\:}21x^2-13x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=-6x+5

Therefore,

\frac{21x^2-13x+5}{3x-1}=7x+\frac{-6x+5}{3x-1}\\\\\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+7x+\frac{-6x+5}{3x-1}

\mathrm{Divide}\:\frac{-6x+5}{3x-1}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{-6x}{3x}=-2\\\\\mathrm{Quotient}=-2

\mathrm{Multiply\:}3x-1\mathrm{\:by\:}-2:\:-6x+2\\\mathrm{Subtract\:}-6x+2\mathrm{\:from\:}-6x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=3

Therefore,

\frac{-6x+5}{3x-1}=-2+\frac{3}{3x-1}\\\\\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+7x-2+\frac{3}{3x-1}

6 0
4 years ago
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