Answer:
B. No.
Step-by-step explanation:
We have been given 3 side lengths 7 ft, 12 ft, 17 ft. We are asked to determine, whether the given set of lengths can form a right triangle or not.
We will use Pythagoras theorem to solve our given problem, which states that the square of hypotenuse of a right triangle is equal to the sum of squares of two legs of right triangle.
![17^2=12^2+7^2](https://tex.z-dn.net/?f=17%5E2%3D12%5E2%2B7%5E2)
![289=144+49](https://tex.z-dn.net/?f=289%3D144%2B49)
![289>193](https://tex.z-dn.net/?f=289%3E193)
Since the sum of squares of both legs is less than square of hypotenuse, therefore, the given set of lengths can not be the side lengths of a right triangle.
Answer:
42 square inches
Step-by-step explanation:
Area of a triangle = (1/2)(base)(height)
A = 1/2(12)(7)
A = 1/2(84)
A = 42
Answer:
$4,800
Step-by-step explanation:
The maximum contribution for traditional IRA in 2019 = $6000
Given that;
karen has a salary of $33,000 and rental income of $33,000; then total income = $66,000
AGI phase-out range for traditional IRA contributions for a single taxpayer who is an active plan participant is $64,000 – $74,000.
PhaseOut can be calculated as: ![\frac{66,000-64000}{74,000-64,000} *6000](https://tex.z-dn.net/?f=%5Cfrac%7B66%2C000-64000%7D%7B74%2C000-64%2C000%7D%20%2A6000)
= ![\frac{2000}{10000} *6000](https://tex.z-dn.net/?f=%5Cfrac%7B2000%7D%7B10000%7D%20%2A6000)
= 0.2 * 6000
= 1200
Therefore, the maximum amount that Karen may deduct for contributions to her traditional IRA for 2019 = The maximum contribution for traditional IRA in 2019 - PhaseOut
= $6000 - $1,200
= $4,800
Answer:
(-8,5)
Step-by-step explanation:
The output corresponds to the y-value of an ordered pair while the input corresponds to the x-value of an order pair.
An ordered pair is written like this (x,y)
Answer: OPTION A
Step-by-step explanation:
The equation of the line in slope-intercept form is:
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
Where m is the slope and b the y-intercept.
Solve for y from each equation:
![\left \{ {{5y=-4x+7} \atop {10y=-8x+14}} \right.\\\\\left \{ {{y=\frac{-4}{5}x+\frac{7}{5}} \atop {y=\frac{-8}{10}x+\frac{14}{10}}} \right.\\\\\left \{ {{y=\frac{-4}{5}x+\frac{7}{5}} \atop {y=\frac{-4}{5}x+\frac{7}{5}}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B5y%3D-4x%2B7%7D%20%5Catop%20%7B10y%3D-8x%2B14%7D%7D%20%5Cright.%5C%5C%5C%5C%5Cleft%20%5C%7B%20%7B%7By%3D%5Cfrac%7B-4%7D%7B5%7Dx%2B%5Cfrac%7B7%7D%7B5%7D%7D%20%5Catop%20%7By%3D%5Cfrac%7B-8%7D%7B10%7Dx%2B%5Cfrac%7B14%7D%7B10%7D%7D%7D%20%5Cright.%5C%5C%5C%5C%5Cleft%20%5C%7B%20%7B%7By%3D%5Cfrac%7B-4%7D%7B5%7Dx%2B%5Cfrac%7B7%7D%7B5%7D%7D%20%5Catop%20%7By%3D%5Cfrac%7B-4%7D%7B5%7Dx%2B%5Cfrac%7B7%7D%7B5%7D%7D%7D%20%5Cright.)
As you can see the slope and the y-intercept of each equation are equal, this means that both are the exact same line. Therefore, you can conclude that the system has infinitely many solutions.