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Aneli [31]
3 years ago
11

Introduction to the triangle midsegment theorem‼️Can someone help me with this problem ❓

Mathematics
1 answer:
Nady [450]3 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

Given a line joining the midpoints of 2 sides of a triangle, the midsegment

Then the midsegment is parallel to and half the measure of the third side of the triangle, thus

XY = 2KL = 2 × 42 = 84

Since J is the midpoint of XY, then

JY = 0.5XY = 0.5 × 84 = 42 and

JK = 0.5YZ = 0.5 × 76 = 38

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5. Find the discriminant of 15x2 = 4x – 1 and describe
diamong [38]

Answer:

a) The discriminant of the equation =  - 44

b)The nature of the roots will be imaginary.

c) x = \frac{2 +\sqrt{11} i}{15}  or, x = \frac{2 - \sqrt{11} i}{15}

Step-by-step explanation:

Here, the given expression is 15x^{2}  = 4x -1

or, 15x^{2}  -  4x + 1   = 0

Now the discriminant (D) of a quadratic equation ax^{2}  +b x + c   = 0

D = b^{2}   -  4ac  = (-4)^{2}  -  4(15) (1)  = 16 - (60) = -44

Hence, the discriminant of the equation =  - 44

As D< 0, so the roots will be imaginary.

Now,by quadratic formula : x = \frac{-b \pm \sqrt{b^{2}  - 4ac} }{2a}

So, here x = \frac{-(-4) \pm \sqrt{D} }{2a}  = \frac{4 \pm \sqrt{(-44 )} }{30}

So, either x = \frac{4 + \sqrt{(-44 )} }{30} or, x =  \frac{4 - \sqrt{(-44 )} }{30}

or, x = \frac{2 +\sqrt{11} i}{15}  or, x = \frac{2 - \sqrt{11} i}{15}

8 0
3 years ago
A percent is a ratio of the ________ to the _______.
riadik2000 [5.3K]

Answer:

A percent is a ratio of the parts to the whole

Step-by-step explanation:

If you had 30/100, 30% as a percentage, 30 would be the parts and 100 would be the whole

7 0
3 years ago
Can someone solve this question PLEASE?? check the attached pic
Talja [164]

Answer:

someone ur so ugly

Step-by-step explanation:

5 0
3 years ago
Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]
alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)²  = 3√6

Therefore, the unit vector is

\frac{[7,2,-1]}{3\sqrt{6} }

or

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

In conclusion, the two unit vectors are;

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

and

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

<em>Hope this helps!</em>

7 0
3 years ago
At a carnival, the cost to play a bowling game is $2, plus $0.40 for each attempt to knock over the pins. Which inequality would
liq [111]

Answer:

a

Step-by-step explanation:

7 0
3 years ago
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