Answer:
x = 48.40'
y = 33.01'
Step-by-step explanation:
These are all segments of a right triangle, so the Pythagorean theorem applies.
x^2 = 46.07^2 +14.82^2 = 2342.0773
x = √2342.0773 ≈ 48.40
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x^2 = 35.39^2 +y^2
y^2 = 2342.0773 -35.39^2 = 1089.6252
y = √1089.6252 ≈ 33.01
Units are feet for both numbers:
x = 48.40'
y = 33.01'
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You always start any question by reading the whole question, identifying the (relevant) given information, and understanding what the question is asking for. You then make an assessment of what you know about the relationships between the given information and what is asked. Finally, you develop a strategy for getting from what you know to what you need to know.
Here, you're given segment lengths of right triangles, and you're asked for other segment lengths. This is not a trig problem--no angles are involved. It is a straight Pythagorean theorem problem. To make use of that theorem, or any formula, you need to have only one unknown. So, you can't start by solving the bottom triangle; you have to start with the upper one where there is only the unknown side x.
After you find x, then you have two known sides in the bottom triangle, so you can find y.
Here, x is an intermediate value in the computation of y. You do NOT use the rounded answer to the question when you're computing y. Rather, you use the full calculator precision for x, so that y can have best accuracy. Only at the end do you round to hundredths.
3)
The first term is 35 and it keeps getting subtracted by 30 to get the other terms so this is the common difference, so we can write:
![\sf a_n = a_{n-1}-30](https://tex.z-dn.net/?f=%5Csf%20a_n%20%3D%20a_%7Bn-1%7D-30)
![\sf a_1=35](https://tex.z-dn.net/?f=%5Csf%20a_1%3D35)
4)
The first term is 5 and it keeps getting subtracted by 8 to get the other terms so this is the common difference, so we can write:
![\sf a_n = a_{n-1}-8](https://tex.z-dn.net/?f=%5Csf%20a_n%20%3D%20a_%7Bn-1%7D-8)
![\sf a_1=5](https://tex.z-dn.net/?f=%5Csf%20a_1%3D5)
5)
The first term is -29 and it keeps getting added by 200 to get the other terms so this is the common difference, so we can write:
![\sf a_n = a_{n-1}+200](https://tex.z-dn.net/?f=%5Csf%20a_n%20%3D%20a_%7Bn-1%7D%2B200)
![\sf a_1=-29](https://tex.z-dn.net/?f=%5Csf%20a_1%3D-29)
6)
The first term is 11 and it keeps getting added by 100 to get the other terms so this is the common difference, so we can write:
![\sf a_n = a_{n-1}+100](https://tex.z-dn.net/?f=%5Csf%20a_n%20%3D%20a_%7Bn-1%7D%2B100)
![\sf a_1=11](https://tex.z-dn.net/?f=%5Csf%20a_1%3D11)
7)
The first term is 37 and it keeps getting added by 30 to get the other terms so this is the common difference, so we can write:
![\sf a_n = a_{n-1}+30](https://tex.z-dn.net/?f=%5Csf%20a_n%20%3D%20a_%7Bn-1%7D%2B30)
![\sf a_1=37](https://tex.z-dn.net/?f=%5Csf%20a_1%3D37)
8)
The first term is 20 and it keeps getting subtracted by 6 to get the other terms so this is the common difference, so we can write:
![\sf a_n = a_{n-1}-6](https://tex.z-dn.net/?f=%5Csf%20a_n%20%3D%20a_%7Bn-1%7D-6)
First find the dimensions of the triangle by using the Pythagorean’s theorem.
a^2+b^2=c^2
*c is always the hypotenuse
24^2+b^2=25^2
576+b^2=625
625-576=b^2
49=b^2
Square root of 49= +or- 7b=+-7
*negative numbers are not often used in length for sides therefore we will be using the positive number.
Tan=opposite/adjacent= 7/24
Sin=opp/hypotenuse= 7/25
Cos = adjacent/hypo= 24/25
Cot=Tan^-1(inverse)= 24/7
Csc = Sin^-1= 25/7
Sec= Cos^-1= 25/24
134.15 the tax price is 9.65 hopes this helps