Answer:![\frac{4}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B5%7D)
Step-by-step explanation:
the fraction is
right, but notice that the numerator and denominator are both divisible by 16.
64 divided by 16 is 4,
80 divided by 16 is 5.
![\frac{4}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B5%7D)
D. When you substitute for the variables the equation becomes -1-5×6=-31. First, you multiply -5 by 6 to get -30. This makes the equation -1-30=-31. -1-30 is -31 so your equation is now -31=-31.
Answer:
1&2 are linear angles, and the other two are adjacent angles!
Step-by-step explanation:
Answer:
![m\angle QSN=65^\circ](https://tex.z-dn.net/?f=m%5Cangle%20QSN%3D65%5E%5Ccirc)
Step-by-step explanation:
In the given figure, PQRS is a rhombus and SRM is an equilateral triangle.
We are also given that SN⊥RM and that ∠PRS = 55°.
And we want to find the measure of ∠QSN.
Remember that since PQRS is a rhombus, the angles formed by its diagonals are right angles. Let the intersection point of the diagonals be K. Therefore:
![m\angle RKS=90^\circ](https://tex.z-dn.net/?f=m%5Cangle%20RKS%3D90%5E%5Ccirc)
Now, RKS is also a triangle. The interior angles of all triangles must be 180. Thus:
![m\angle RKS+m\angle KSR+m\angle SRK=180](https://tex.z-dn.net/?f=m%5Cangle%20RKS%2Bm%5Cangle%20KSR%2Bm%5Cangle%20SRK%3D180)
Substitute in known values:
![90+55+m\angle KSR=180](https://tex.z-dn.net/?f=90%2B55%2Bm%5Cangle%20KSR%3D180)
Solve for ∠KSR:
![m\angle KSR+145=180\Rightarrow m\angle KSR=35^\circ](https://tex.z-dn.net/?f=m%5Cangle%20KSR%2B145%3D180%5CRightarrow%20m%5Cangle%20KSR%3D35%5E%5Ccirc)
Since SRM is an equilateral triangle, this means that:
![m\angle SRM=m\angle RMS=m\angle MSR=60^\circ](https://tex.z-dn.net/?f=m%5Cangle%20SRM%3Dm%5Cangle%20RMS%3Dm%5Cangle%20MSR%3D60%5E%5Ccirc)
Note that RNS is also a triangle. Therefore:
![m\angle SRM+m\angle RNS+m\angle NSR=180](https://tex.z-dn.net/?f=m%5Cangle%20SRM%2Bm%5Cangle%20RNS%2Bm%5Cangle%20NSR%3D180)
Substitute in known values:
![60+90+m\angle NSR=180](https://tex.z-dn.net/?f=60%2B90%2Bm%5Cangle%20NSR%3D180)
So:
![m\angle NSR+150=180\Rightarrow m\angle NSR=30^\circ](https://tex.z-dn.net/?f=m%5Cangle%20NSR%2B150%3D180%5CRightarrow%20m%5Cangle%20NSR%3D30%5E%5Ccirc)
∠QSN is the addition of the two angles:
![m\angle QSN=m\angle KSR+m\angle NSR](https://tex.z-dn.net/?f=m%5Cangle%20QSN%3Dm%5Cangle%20KSR%2Bm%5Cangle%20NSR)
Therefore:
![m\angle QSN=35+30=65^\circ](https://tex.z-dn.net/?f=m%5Cangle%20QSN%3D35%2B30%3D65%5E%5Ccirc)