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Delvig [45]
4 years ago
5

What is the measure of angle Y?

Mathematics
2 answers:
MrRa [10]4 years ago
4 0
24 is the answer because if you try any other answer it would be bigger than 180.
avanturin [10]4 years ago
4 0

Answer:

I just took  the test the answer is indeed 24

Step-by-step explanation:

You might be interested in
To the nearest tenth, find the perimeter of ABC with vertices A (-2,-2) B (0,5) and C (3,1)
Pavlova-9 [17]

the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.


\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}


\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill

5 0
3 years ago
Can someone answer questions 20-24 I need it for my math homework
navik [9.2K]
You know you have to draw the triangles right?
3 0
2 years ago
How do u get the answer for 42x3/6
sasho [114]
42*3/6 is equal to 42*1/2, or 21.
8 0
4 years ago
The temperature, T, of a given mass of gas varies inversely with its Volume,V. The temperature of 98 cm ³ of a certain gas is 28
Naily [24]

Answer:

T₂ = 421.4 K

Step-by-step explanation:

The temperature, T, of a given mass of gas varies inversely with its Volume,V.

T\propto \dfrac{1}{V}\\\\V_1T_1=V_2T_2

We have,

V₁ = 98 cm³, T₁ = 28° C = (28+273) K = 301 K

V₂ = 70 cm₃

We need to find the new temperature of the gas. Using the above relation to find T₂.

T_2=\dfrac{V_1T_1}{V_2}\\\\T_2=\dfrac{98\times 301}{70}\\\\T_2=421.4\ K

Hence, the required temperature of the gas is 421.4 K.

4 0
2 years ago
Please help????!!!!!
rjkz [21]

Answer:

y =8

Step-by-step explanation:

y^{2} = 16*4

4 0
3 years ago
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