Isolate the x. Note the equal sign. What you do to one side, you do to the other.
First, subtract 2 from both sides
2 (-2) - x = (5/8) - 2
-x = 5/8 - 2
Make sure they have the same denominator. Remember that what you multiply to the denominator, you multiply the numerator:
-x = 5/8 - 16/8
-x = - 11/8
Isolate the x. Divide -1 from both sides
(-x)/-1 = (-11/8)/-1
x = 11/8
11/8 is your answer for x
<em>~Rise Above the Ordinary</em>
Answer:
The numbers slowly decrease by 4 so we can determine that the 40th sequence is -70.
Step-by-step explanation:
Answer:
506.7 cm²
Step-by-step explanation:
The amount of cardboard needed would be the surface area of the cone,
S= pi(r)(l+r) where r is the radius and l is the slant height.
You can find l using the Pythagorean theorem. l²=r²+h² which is l²=6²+20² and you get l = √436.
Plug r and l into the equation: S = 6pi(√436 + 6) = 506.7 cm²
Let me know if you have any questions about my steps!
Answer:
11 quarters
Step-by-step explanation:
If Thomas ignores the extra 3 dimes, he can arrange his coins in groups of 3 dimes and 1 quarter, each group valued at 55¢. It takes
... 6.05/0.55 = 11
groups to bring the total value to the $6.05 remaining after ignoring the extra dimes.
Thomas has 11 quarters and 36 dimes.
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]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20x_%7B1%7D%26y_%7B1%7D%261%5C%5Cx_%7B2%7D%26y_%7B2%7D%261%5C%5Cx_%7B3%7D%26y_%7B3%7D%261%5Cend%7Barray%7D%5Cright%5D%20)
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