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Svet_ta [14]
3 years ago
13

can someone help? I'm getting ready to slit someone's throat, I'm so irritated. I have 7 of these to do. ​

Mathematics
1 answer:
Brums [2.3K]3 years ago
8 0
97.8ft^2 but if you round it to the nearest 10 it will be 100ft^2
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100 points, PLEASE HELP MEEE
masha68 [24]

Answer:

b = 3a

Step-by-step explanation:

a = < - 1, 3 >

b = < - 3, 9 >

Thus

b = 3a

7 0
2 years ago
On a coordinate plane, a square and a point are shown. The square has points R prime (negative 8, 1), S prime (negative 4, 1), T
VladimirAG [237]

Answer:

(–1, –6)

Step-by-step explanation:

Given that S(3, –5) was translated to S'(–4, 1), the transformation rule is (x-7, y+6). Then, the coordinates of square RSTU are:

R'(–8, 1) -> (-8+7, 1-6) -> R(-1, -5)

S'(–4, 1) -> (-4+7, 1-6) -> S(3, -5)

T'(–4, –3) -> (-4+7, -3-6) -> T(3, -9)

U'(–8, –3) -> (-8+7, -3-6) -> U(-1, -9)

The point (–1, –6) lies on the segment RU

7 0
3 years ago
Read 2 more answers
1) Which variable did you plot on the x-axis, and which variable did you plot on the y-axis?
Liula [17]

Answer:

Step-by-step explanation:

This question is too difficult

3 0
3 years ago
Please help i will igve brainliest
adelina 88 [10]
3. you cannot simplify √30
4 0
3 years ago
Match each set of vertices with the type of triangle they form.
Andrew [12]

Answer:  The calculations are done below.


Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

4 0
3 years ago
Read 2 more answers
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