49 is the answer to ur question
Answer:
-2 > -6 . . . or . . . R > S
Step-by-step explanation:
If you plot the temperatures of the two cities on a number line, you see that the temperature for City R is plotted to the right of the temperature for City S. This means the temperature of City S is less than that of City R.
If your inequality relates cities:
S < R or R > S
If your inequality relates temperatures:
-6 < -2 or -2 > -6
I need more info to answer this question
Answer:
Step-by-step explanation:
Lateral surface area of the triangular prism = Perimeter of the triangular base × Height
By applying Pythagoras theorem in ΔABC,
AC² = AB² + BC²
(34)² = (16)² + BC²
BC = ![\sqrt{1156-256}](https://tex.z-dn.net/?f=%5Csqrt%7B1156-256%7D)
= ![\sqrt{900}](https://tex.z-dn.net/?f=%5Csqrt%7B900%7D)
= 30 in.
Perimeter of the triangular base = AB + BC + AC
= 16 + 30 + 34
= 80 in
Lateral surface area = 80 × 22
= 1760 in²
Total Surface area = Lateral surface area + 2(Surface area of the triangular base)
Surface area of the triangular base = ![\frac{1}{2}(\text{Base})(\text{Height})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Ctext%7BBase%7D%29%28%5Ctext%7BHeight%7D%29)
= ![\frac{1}{2}(30)(16)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2830%29%2816%29)
= 240 in²
Total surface area = 1760 + 2(240)
= 1760 + 480
= 2240 in²
Volume = Area of the triangular base × Height
= 240 × 20
= 4800 in³
Answer:
The area of the triangle is ![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)
Step-by-step explanation:
Given:
Coordinates D (0, 0), E (1, 1)
Angle ∠DEF = 60°
△DEF is a Right triangle
To Find:
The area of the triangle
Solution:
The area of the triangle is = ![\frac{1}{2}(base \times height)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28base%20%5Ctimes%20height%29)
Here the base is Distance between D and E
calculation the distance using the distance formula, we get
DE = ![\sqrt{(0-1)^2 + (0-1)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%280-1%29%5E2%20%2B%20%280-1%29%5E2%7D)
DE =![\sqrt{(-1) ^2 + (-1)^2](https://tex.z-dn.net/?f=%5Csqrt%7B%28-1%29%20%5E2%20%2B%20%28-1%29%5E2)
DE = ![\sqrt{1+1}](https://tex.z-dn.net/?f=%5Csqrt%7B1%2B1%7D)
DE = ![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D)
Base = ![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D)
Height is DF
DF =![tan(60^{\circ}) \times DE](https://tex.z-dn.net/?f=tan%2860%5E%7B%5Ccirc%7D%29%20%5Ctimes%20DE)
DF = ![\sqrt{3} \times DE](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20%5Ctimes%20DE)
DF = ![\sqrt{3} \times\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20%5Ctimes%5Csqrt%7B2%7D)
Now, the area of the triangle is
= ![\frac{1}{2}({\sqrt{2})(\sqrt{3} \times \sqrt{2})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%7B%5Csqrt%7B2%7D%29%28%5Csqrt%7B3%7D%20%5Ctimes%20%5Csqrt%7B2%7D%29)
=![\frac{1}{2}({\sqrt{2})(\sqrt{3} \sqrt{2})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%7B%5Csqrt%7B2%7D%29%28%5Csqrt%7B3%7D%20%5Csqrt%7B2%7D%29)
=![\frac{1}{2}(2\sqrt{3} )](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%282%5Csqrt%7B3%7D%20%29)
=![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)