Answer: 5 5/12 miles
Step-by-step explanation:
The following are the distance covered by Jada for the week.
Monday = 1 1/3 miles
Tuesday = 5/6 miles
Wednesday = 0
Thursday = 2 3/4 miles
Friday = 1 1/2 miles
Total distance for the week will be:
= 1 1/3 + 5/6 + 2 3/4 + 1 1/2
The lowest common multiple of 3,6,4 and 2 is 12. The fractions will then be
1 4/12 + 10/12 + 2 9/12 + 6/12
= 3 29/12
= 3 + 2 5/12
= 5 5/12
Jada ran 5 5/12 miles this week
Answer:
Step-by-step explanation:
square- 48*32=1536
triangle- .5*12*48= 288
9514 1404 393
Answer:
(d) ∠H ≅ ∠J
Step-by-step explanation:
We already know that ∠G is congruent to itself. If we show (by translation or other means) that ∠I ≅ ∠K, then we know that ΔGHI ~ ΔGJK. The third angle in each triangle will be congruent, too.
∠H ≅ ∠J
_____
The problem is concerned with angles, so the first two answer choices are irrelevant. If two angles are shown congruent, the triangles are congruent by AA similarity, so the third answer choice is incorrect.
Answer:
Consequently, any trigonometric identity can be written in many ways. To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation.
Step-by-step explanation:
Answer:
it must also have the root : - 6i
Step-by-step explanation:
If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.
This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:
![(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2](https://tex.z-dn.net/?f=%28x-%28a%2Bbi%29%29%2A%28x-%28a-bi%29%29%3D%5C%5C%28x-a-bi%29%2A%28x-a%2Bbi%29%3D%5C%5C%28%5Bx-a%5D-bi%29%2A%28%5Bx-a%5D%2Bbi%29%3D%5C%5C%28x-a%29%5E2-%28bi%29%5E2%3D%5C%5C%28x-a%29%5E2-b%5E2%28-1%29%3D%5C%5C%28x-a%29%5E2%2Bb%5E2)
where the imaginary unit has disappeared, making the expression real.
So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)
Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.
