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3 years ago

I have absolutely no idea what’s going on

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

8 0

Answer:

The hypotenuse( longer side) of the triangle = $c^{2} = a^{2} +b^{2} \\Therfore, c = \sqrt a^{2} +b{2}\\ \\$

Step-by-step explanation:

Here, side =c

Therfore, area = c *c

That is $a^{2} +b{2}$ = 50 $^{2} + 35^{2}$ = 3250 = 57 units sq

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If the 50th and 51st terms of an arithmetic sequence are 140 and 142, find the 1st term

morpeh [17]

Answer:

The first term is 42.

Step-by-step explanation:

The common difference is 142 - 140 = 2 so we have

50th term = a1 + 2(50 -1) = 140 where a1 is the first term

a1 + 98 = 140

a1 = 140 - 98 = 42.

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3 years ago

What is the equation of the line through (3,-1), (4,7)

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Answer:

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3 years ago

Can i get help on 18 & 22 please

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3 years ago

What is the equation of the line passing through ( -5, -3 ) and perpendicular

Artist 52 [7]

Answer:

y = 7/5x + 4

Step-by-step explanation:

use the slope from the equation -5/7 and take the negative reciprocal to get the perpendicular slope = 7/5. Then use the equation y=mx+b. Plug in x and y from the point given and the new slope and solve for b.

(-3) = (7/5)(-5) + b, (-3) = -7 + b, add 7 to both sides. b = 4. Rewrite the equation now to be y = 7/5x + 4

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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

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