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Anarel [89]
3 years ago
5

I need help anyone can help me

Mathematics
1 answer:
REY [17]3 years ago
5 0
I found the answer at another website plus wanna get exposed little boys yall aint girls
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Anybody please help me with this.... I have a very important grade relying on this
Anon25 [30]
Five is 11 and negative 11 because double negative equals a positive
3 0
3 years ago
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Find the area of the circle r=6 ft 113.04 ft2, 9.42 ft2, 28.26ft2, 37.68 ft2, 18.84 ft2
nadya68 [22]

Answer:

113.04ft^2

Step-by-step explanation:

Like before A=\pi  r^2\\=\pi 6^2\\=113.04ft^2

6 0
3 years ago
Eli has 8 black pens and 5 in his desk drawer. He also has 3 yellow highlighters, 4 green highlighters, and 5 pink highlighters
34kurt

Answer:

The answer is 15/52. (Please make me as brainliest (: )

7 0
3 years ago
What is the slope of a line that is perpendicular to the line shown?
ladessa [460]

Answer:

Step-by-step explanation:

(0,2)  (3,0)

(0-2)/(3-0)= -2/3

the perpendicular slope is 3/2

6 0
3 years ago
In the graph below, Point A represents Owen's house, Point B represents David's house and Point C represents the school. Who liv
ycow [4]

Answer:

• David

,

• 4 miles

Explanation:

In the graph:

The given locations are:

• Owen's House, A(11,3)

,

• David's House, B(15,13)

,

• School, C(3,18)

We determine both Owen's and David's distance from the school using the distance formula.

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

Owen's distance from school (AC)

\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}

David's distance from school (BC)

\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}

We see from the calculations that David lives closer to the school, and by 4 miles.

The graph below is attached for further understanding:

5 0
1 year ago
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