First, you should graph the points. For the first number, called the X-Axis, you should to the right or left, and for the second number, called the Y-Axis, you should go up or down.
To find the distance between Point A and Point C, you should simply just count the number of intersections between them (4).
Angle B is a right angle because if the triangle is bisected at B, it will leave a right angle on either side. Therefore, to label it, you should simply just draw a line through Point B all of the way to line (A,C).
The type of triangle you have drawn is an isosceles, because it has 2 equal angles and 2 equal sides.
We know both of the sides that are unknown will be the same because the triangle is bilateral. Then, we can use the bisection we made earlier to solve for the unknown sides using Pythagorean Theorem. Since earlier, we know the entire bottom is 4, we know half of the bottom is 2. We can also see that the height of the triangle is 2. We then plug those numbers into the Pythagorean Theorem (A^2*B^2=C^2) which makes the value of C^2=16. We then take the square root of C^2 and 16 to see that both unknown sides are 4.
Answer:
You'd earn $825 in interest.
Step-by-step explanation:
The appropriate formula for simple interest is i = p·r·t, where p is the principal, r is the interest rate as a decimal fraction, and t is the time in years. Because 8 months constitutes 2/3 year, we have:
i = $45,000·(0.0275)(2/3) = $825.
You'd earn $825 in interest.
Let the hall be A=1. For 3 hours 16 people can decorate the hall. Then P1=A\(t1.number people1) P=1/48. That is for 1 person. You have 24 people to decorate the same hall. Their P2 will be (1/48).24=1/2. t2=A/P2=2 hours
(-3,-5)(2,4)
slope(m) = (4 - (-5) / (2 - (-3) = (4 + 5) / (2 + 3) = 9/5
y - y1 = m(x - x1)
slope(m) = 9/5
(2,4)...x1 = 2 and y1 = 4
sub
y - 4 = 9/5(x - 1)....this is the correct answer
the student figured the slope wrong....it is 9/5...not -9/5
Answer:
The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.
Step-by-step explanation:
This is a problem of optimization.
We have to minimize the time it takes for the lifeguard to reach the child.
The time can be calculated by dividing the distance by the speed for each section.
The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.
Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:
Then, the time (speed divided by distance) is:
To optimize this function we have to derive and equal to zero:
![\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\](https://tex.z-dn.net/?f=%5Cdfrac%7Bdt%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B4%7D%2B%5Cdfrac%7B1%7D%7B1.1%7D%28%5Cdfrac%7B1%7D%7B2%7D%29%5Cdfrac%7B2x-120%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%5C%5C%5C%5C%5C%5C%5Cdfrac%7Bdt%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B4%7D%20%2B%5Cdfrac%7B1%7D%7B1.1%7D%20%5Cdfrac%7Bx-60%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%3D0%5C%5C%5C%5C%5C%5C%20%20%5Cdfrac%7Bx-60%7D%7B%5Csqrt%7Bx%5E2-120x%2B5200%7D%7D%20%3D%5Cdfrac%7B1.1%7D%7B4%7D%3D%5Cdfrac%7B2%7D%7B7%7D%5C%5C%5C%5C%5C%5C%20x-60%3D%5Cdfrac%7B2%7D%7B7%7D%5Csqrt%7Bx%5E2-120x%2B5200%7D%5C%5C%5C%5C%5C%5C%28x-60%29%5E2%3D%5Cdfrac%7B2%5E2%7D%7B7%5E2%7D%28x%5E2-120x%2B5200%29%5C%5C%5C%5C%5C%5C%28x-60%29%5E2%3D%5Cdfrac%7B4%7D%7B49%7D%5B%28x-60%29%5E2%2B40%5E2%5D%5C%5C%5C%5C%5C%5C%281-4%2F49%29%28x-60%29%5E2%3D4%2A40%5E2%2F49%3D6400%2F49%5C%5C%5C%5C%2845%2F49%29%28x-60%29%5E2%3D6400%2F49%5C%5C%5C%5C45%28x-60%29%5E2%3D6400%5C%5C%5C%5C)
As 
, the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.
