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Usimov [2.4K]
3 years ago
11

PLEASE HELP!!!!!!!!!! Solve this system of linear equations. Separate

Mathematics
1 answer:
g100num [7]3 years ago
7 0

Answer:

14.5,8.75

Step-by-step explanation:

Given equations:

             -3x = -26 -2y  ----------- i

             -11x  = 90 - 2y  --------- ii

To solve this problem, simply subtract equation ii from i;

        [-3x -(-11x)] = {-26 - 90] -[-2y - (-2y)]

             8x  = 116

                x = 14.5;

Then substitute x  = 14.5 into i to solve for y;

             -3(14.5) = -26 - 2y

               -43.5 = -26 -2y

               -43.5 + 26 = -2y

                   -17.5  = -2y

                       y = 8.75

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The sides of ∆ABC inscribed in a circle are equidistant from the center. Tricia says that ∆ABC must be equilateral. Which explai
lisabon 2012 [21]

Answer:

Tricia is correct because the sides of the triangle are congruent since they are  equidistant from the center of the circle.

Step-by-step explanation:

Given that the sides of ΔABC inscribed in a circle are equidistant from the center. This implies the triangle is drawn within the circumference of the circle with a center.

Bisecting the three sides of the triangle shows that the three bisectors intersect at a point which is the center of the circle, and the lines are equidistant to this point. This proves that the length of the sides of the triangle are equal, so that the triangle is an equilateral triangle. Therefore, Tricia's statement is correct.

3 0
3 years ago
Help I have 10 mins.
Paul [167]

Answer:

\sqrt{0.5}

Step-by-step explanation:

cosx=adj/hyp

adj= \sqrt{2}/2

hyp= r

cos(45)= (\sqrt{2}/2)/r

\sqrt{2}/2 = (\sqrt{2}/2)/r

   multiply both sides by r

(\sqrt{2}/2)r = \sqrt{2}/2

    divide both sides by \sqrt{2}/2

r = 1

pythagorean theorem: a^{2} + b^{2} = c^{2}

a = \sqrt{2}/2

b = y

c = r = 1

(\sqrt{2}/2)^{2} + y^{2} = 1^{2}

    (\sqrt{2}/2)^{2} = 2/4 = 0.5

0.5 + y^{2} = 1

    subtract 0.5 from 1

y^{2} = 0.5

\sqrt{y^{2}} = \sqrt{0.5}

y = \sqrt{0.5}

7 0
3 years ago
In the rolling of two fair dice calculate the following: P(Sum of the two dice is 7) = ______
vesna_86 [32]

Answer:

P(Sum of the two dice is 7) = 6/36

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that:

A fair dice can have any value between 1 and 6 with equal probability. There are two fair dices, so we have the following possible outcomes.

Possible outcomes

(first rolling, second rolling)

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)

(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)

(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)

(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)

(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)

(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

There are 36 possible outcomes.

Desired outcomes

Sum is 7, so

(1,6), (6,1), (5,2), (2,5), (3,4), (4,3).

There are 6 desired outcomes, that is, the number of outcomes in which the sum of the two dice is 7.

Answer

P(Sum of the two dice is 7) = 6/36

7 0
4 years ago
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