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RoseWind [281]
2 years ago
8

Can some one help me with this

Mathematics
2 answers:
N76 [4]2 years ago
7 0
I think this is how you solve this and the answer  <span><span><span>0.1+30</span>+8</span>−12</span>+<span>14</span><span>=<span>26.35

</span></span>
cestrela7 [59]2 years ago
6 0
Plug in the numbers for the variables
0.1(30)+8-12(1/4)
3+8-3
8
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Aneli [31]
What is the complete question?
5 0
3 years ago
A survey taken during class revealed 30% of the students in a class bring their lunch to school. If the probability that a rando
ohaa [14]
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</span>P(boy |lunch) = P(boy and lunch)/P(lunch) = 0.40
P(boy and lunch) = P(lunch)*0.4 = 0.3*0.4 = 0.12
8 0
3 years ago
You’re help Is greatly appreciated!! I will mark BRAINLIEST as well
Akimi4 [234]

Answer:

1. -4/7

2. 4/3

3. x^2 + 16x + 63

4. x^2 + 19x + 90

Step-by-step explanation:

1. (1, 8) and (8, 4)

slope = m = (8 - 4)/(1 - 8) = -4/7

2. (2, 4) and (5, 8)

slope = m = (8 - 4)/(5 - 2) = 4/3

3. (x + 9)(x + 7) =

= x^2 + 7x + 9x + 63

= x^2 + 16x + 63

4. (x + 10)(x + 9) =

= x^2 + 9x + 10x + 90

= x^2 + 19x + 90

6 0
2 years ago
If A(-5,7), B(-4,-5), C(-1,-6) and D(4,5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.
MrMuchimi
After plotting the quadrilateral in a Cartesian plane, you can see that it is not a particular quadrilateral. Hence, you need to divide it into two triangles. Let's take ABC and ADC.

The area of a triangle with vertices known is  given by the matrix
M = \left[\begin{array}{ccc} x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right]

Area = 1/2· | det(M) |
        = 1/2· | x₁·y₂ - x₂·y₁ + x₂·y₃ - x₃·y₂ + x₃·y₁ - x₁·y₃ |
        = 1/2· | x₁·(y₂ - y₃) + x₂·(y₃ - y₁) + x₃·(y₁ - y₂) |

Therefore, the area of ABC will be:
A(ABC) = 1/2· | (-5)·(-5 - (-6)) + (-4)·(-6 - 7) + (-1)·(7 - (-5)) |
             = 1/2· | -5·(1) - 4·(-13) - 1·(12) |
             = 1/2 | 35 |
             = 35/2

Similarly, the area of ADC will be:
A(ABC) = 1/2· | (-5)·(5 - (-6)) + (4)·(-6 - 7) + (-1)·(7 - 5) |
             = 1/2· | -5·(11) + 4·(-13) - 1·(2) |
             = 1/2 | -109 |
<span>             = 109/2</span>

The total area of the quadrilateral will be the sum of the areas of the two triangles:

A(ABCD) = A(ABC) + A(ADC) 
               = 35/2 + 109/2
               = 72
8 0
3 years ago
The area of a rectangular invitation card is 3/10 square foot. Given that the length of the card is 4/5 foot find its perimeter
Andrei [34K]
The perimeter would be 3/8+3/8+4/5+4/5= 2.35 or 2 7/20 square foot.
I hope this helps.
YOU'RE WELCOME :D
8 0
3 years ago
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