Add 3 to both sides so that the equation becomes -2x^2 + 5x + 5 = 0.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -5 ± √(5^2 - 4(-2)(5)) ] / ( 2(-2) )
x = [-5 ± √(25 - (-40) ) ] / ( -4 )
x = [-5 ± √(65) ] / ( -4)
x = [-5 ± sqrt(65) ] / ( -4 )
x = 5/4 ± -sqrt(65)/4
The answers are 5/4 + sqrt(65)/4 and 5/4 - sqrt(65)/4..
Answer:
11 years
Step-by-step explanation:
14 divide 2=7
7 divide 2=3.5
3.5 divide 2 = 175 subtract 1 extra year to turn to 200. So the answer is 11.
Answer:
9408,Is the only number that can divide into the same number to get exactly down to 1.
Step-by-step explanation:
Answer:
30 degrees
Step-by-step explanation:
To figure out theta is using a bit of simple trigonometry.
2sin(theta) = 1
= sin(theta) = 0.5
Using arcsin (sin^-1) we can isolate x
x = arcsin(0.5)
x = 30
Answer:
Therefore they are 734.106 miles apart.
Step-by-step explanation:
Given that ,
Two ships have a harbor together. The angle between two ships is 135°40'. Each of two ships travel 402 miles.
It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.
Let ∠B= 135°40', and AB = 402 miles , BC = 402 miles
Then the distance between the ships = AC
We know
The sum of all angles = 180°
⇒∠A+∠B+∠C=180°
⇒∠A+135°40'+∠C=180°
⇒2∠A= 180°- 135°40' [ since ∠A=∠C]
⇒2∠A=44°60'
⇒∠A= 22°30'
Again we know that,
![\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}](https://tex.z-dn.net/?f=%5Cfrac%7BAB%7D%7Bsin%5Cangle%20C%7D%3D%5Cfrac%7BBC%7D%7Bsin%20%5Cangle%20A%7D%3D%5Cfrac%7BAC%7D%7Bsin%20%5Cangle%20B%7D)
Taking last two ratio,
![\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}](https://tex.z-dn.net/?f=%5Cfrac%7BBC%7D%7Bsin%20%5Cangle%20A%7D%3D%5Cfrac%7BAC%7D%7Bsin%20%5Cangle%20B%7D)
Putting the value of BC , AC ,∠A,∠B
![\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}](https://tex.z-dn.net/?f=%5Cfrac%7B402%7D%7Bsin%2022%5E%5Ccirc30%27%7D%3D%5Cfrac%7BAC%7D%7Bsin%20135%5E%5Ccirc40%27%7D)
![\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}](https://tex.z-dn.net/?f=%5CRightarrow%20AC%3D%5Cfrac%7B402%20%5Ctimes%20sin135%5E%5Ccirc40%27%7D%7Bsin%2022%5E%5Ccirc30%27%7D)
≈734.106 miles
Therefore they are 734.106 miles apart.