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telo118 [61]
3 years ago
13

2+2 Please help I’ve been try this all day long

Mathematics
2 answers:
Alika [10]3 years ago
6 0

Answer:

4

Step-by-step explanation:

LOL

sesenic [268]3 years ago
5 0

Answer:

the answer is 4!!!!!!!!!!!!!!

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Write the equation of the line with a slope of -5 and y-intercept of 12.
Wittaler [7]

Answer:

y=-5x+12

Step-by-step explanation:

Since slope-intercept form is y=mx+b, m is the slope, and b is the y-intercept. We substitute m for -5, since it is the slope, and we substitute 12 for b, since it is the y-intercept. I hope this helps!

7 0
3 years ago
I need some help…
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The total amount after 5 years would be $2,650
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3 years ago
What is k- 1.8= -10.5
Elza [17]
K=8.7

M=2

B=240

X=.2/.3
8 0
3 years ago
How do i do 2x8 and don’t just tell me the answer i need someone to explain pls
Sonja [21]

Answer:

16

Step-by-step explanation:

2 x 8 is just another way of saying what would 8 + 8 be

for example, 5x5 would just 25 but 5+5+5+5+5 would also be 25

Hope this helps, sorry im not a good teacher:)

5 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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