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Kay [80]
4 years ago
14

Sharon and Jacob started at the same place. Jacob walked 3 m north and then 4 m west. Sharon walked 5 m south and 12 m east. How

far apart are Jacob and Sharon now?
A) 5 m
B) 8 m
C) 13 m
D) 18 m
Mathematics
1 answer:
Nesterboy [21]4 years ago
4 0

Answer:

d

Step-by-step explanation:

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Solve attachment EASY
ANTONII [103]

Answer:

x=\frac{-3d-9}{k-d}

Step-by-step explanation:

Step 1: Write out equation

d(-3 + x) = kx + 9

Step 2: Distribute

-3d + dx = kx + 9

Step 3: Subtract dx on both sides

-3d = kx - dx + 9

Step 4: Subtract 9 on both sides

-3d - 9 = kx - dx

Step 5: Factor

-3d - 9 = x(k - d)

Step 6: Divide both sides by k - d

x=\frac{-3d-9}{k-d}

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Molly's Milk Company wants to repaint the barrels in which the milk is stored. The workers need to know how much area they will
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 he surface area of one barrel would be 489.84 
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Solving systems of linear inequalities
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No (4,11) it’s a solution of the system
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Given that Triangle ABC ~ DEF, solve for X.​
Vadim26 [7]

Answer:

x = 7

Step-by-step explanation:

if ABC ~ DEF than AB ~ ED, AC ~ DF, and CB ~ FE

The ratio of AB to ED is given as 5/30 = 1/6

For AC ~ ED the ratio would be same x/42 = 1/6 cross multiply it

6x = 42

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6 0
3 years ago
Find the area of quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21)
sasho [114]

Answer:

The area of quadrilateral ABCD is 139 unit^2.

Step-by-step explanation:

Given:

Quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21).

Now, to find the area.

The coordinates of the quadrilateral are A(1,1), B(7,-3), C(12,2), D(7,21).

So, to find the area we need to bisect the quadrilateral ABCD and get the triangles ABC and ADC and then calculate their areas:

In Δ ABC:

A(x_1,y_1)=(1,1)\:,\:B(x_2,y_2)=(7,-3)\:and\:C(x_3,y_3)=(12,2)

Now, to get the area of triangle ABC:

Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-3-2)+(7)(2-1)+12(1--3)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-5)+(7)(1)+12(4)\right|

On solving we get:

Area\,of\,triangle\,=25.

In Δ ADC:

A(x_1,y_1)=(1,1)\:,\:D(x_2,y_2)=(7,21)\:and\:C(x_3,y_3)=(12,2)

Now, to get the area of triangle ADC:

Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(21-2)+(7)(2-1)+12(1-21)\right|

On solving it by same process as above we get:

Area\,of\,triangle\,=114

Now, to get the area of the quadrilateral we add the areas of the triangles ABC and ADC:

25+114\\=25+114\\=139\ unit^2

Therefore, area of quadrilateral ABCD is 139 unit^2.

4 0
3 years ago
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