What’s this I don’t know

what could be a side length of a right triangle A.) 30 in, 45 in, 50 in B.) 30 in, 40 in,50 in C.) 30 in, 40 in, 60 in D.) 25 in

lions [1.4K]

The pythagorean theorem holds for every right triangle: given the legs $a, b$ and the hypothenuse $c$ , the triangle is right if and only if

$a^2+b^2=c^2$

So, you have to check:

$30^2+45^2=2925\neq 2500 = 50^2$

So the first triangle can't be a right triangle.

$30^2+40^2=2500= 50^2$

So the second triangle is a right triangle.

The third triangle can't be right, because it has the same legs but a different hypothenuse

Finally, we have

$25^2+40^2=2225= 50^2$

So the last triangle can't be a right triangle.

4 0

3 years ago

Ben sold his small online business for $100,000. The purchaser will pay him $20,000 today, then $20,000 every year for the next

kifflom [539]

Answer:

c. 92,598

Step-by-step explanation:

20000 + 20000/1.04 + 20000/(1.04^2) + 20000/(1.04^3) + 20000/(1.04^4)= 20000 + 19230.77 + 18491.12 + 17779.73 + 17096.08

= 92597.

= 92,598

7 0

3 years ago

How do you do 83 divided by 3 with long division

bulgar [2K]

Answer:83÷3=27.6 or 27.667

Step-by-step explanation:

3 0

4 years ago

Which points have x coordinates that are greater than 6?

Mekhanik [1.2K]

B

Locate the point 6 on the x-axis

Any point to the left of 6 has an x-coordinate less than 6 while every point to the right of 6 has an x-coordinate greater than 6

The points to the right of 6 are Z and L

8 0

3 years ago

Fit a trigonometric function of the form f(t)=c0+c1sin(t)+c2cos(t)f(t)=c0+c1sin⁡(t)+c2cos⁡(t) to the data points (0,5.5)(0,5.5),

larisa86 [58]

Answer

$f(t)=-0.2+4.1sin(t)+4cos(t)$

Step-By-Step Explanation

Given the function $f(t)=c_0+c_1sin(t)+c_2cos(t)$ .

For each pair (t, f(t)) in the data points (0,5.5), (π/2,0.5), (π,−2.5), (3π/2,−7.5)

$f(0)=c_0+0c_1+c_2=5.5$ .

$f(\pi /2)=c_0+c_1+0c_2=0.5$ .

$f(\pi)=c_0+0sin(t)-c_2=-2.5$ .

$f(3\pi /2)=c_0-c_1+0c_2=-7.5$ .

Expressing this as a system of linear equations in matrix form AX=B

$\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)\left( \begin{array}{c} c_{0} \\ c_{1} \\ c_{2}\\ \end{array} \right)=\left(\begin{array}{c} 5.5 \\ 0.5 \\ -2.5 \\ -7.5 \end{array} \right)$

Where

$A=\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)$ ,

$B=\left(\begin{array}{c}5.5\\0.5\\-2.5\\-7.5\end{array} \right)$

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)$

To determine the values of X, we use the expression

$X=(A^{T}A)^{-1}A^{T}B$

$A^{T}A= \left(\begin{array}{ccc} 3 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right)$

$(A^{T}A)^{-1}= \left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)$

$A^{T}B=\left(\begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$

Therefore:

$X=\left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)\left( \begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)=\left(\begin{array}{c} -0.2 \\4.1\\4\end{array}\right)$

Therefore, the trigonometric function which fits to the given data is:

$f(t)=-0.2+4.1sin(t)+4cos(t)$

8 0

3 years ago