Answer:
B
Step-by-step explanation:
I Had this
Answer:
86
Step-by-step explanation:
<u>Perimeter of WXY = WSY+WRX+XY</u>
<em>--> WSY = SY x 2</em>
--> WSY = 16 x 2 = 32
<em>Since it is an isosceles triangle, WRX = WSY</em>
--> WRX = 32
<em>--> Draw a straight line from W to XY to divide it into two halves assuming it to be point A. This would form a right angle triangle of WAX.</em>
<em>--> Solve it using the cos theta rule</em>
--> Angle = Angle X = 70°
Hypotenuse = WRX = 32
Adjacent = WA = ?
<em>--> Cos (Angle) = Adjacent/Hypotenuse</em>
Cos (70) = WA/32
WA = 10.9 rounded off to 11
--> WA=AY= 11
--> XY = WA + AY = 11+11 = 22
<em>--> Perimeter = WSY+WRX+XY</em>
Perimeter = 32+32+22
Perimeter = 86
Therefore, the perimeter of WXY is 86.
The exponential function that describes the graph is 
The standard form of an exponential function is expressed as 
a is the y-intercept
(x, y) is the point on the graph
Given the following expression
a = 7
(x, y) = (4, 112)
Substitute the given values into the exponential equation
![y = ab^x\\112=7\cdot b^4\\b^4 = \frac{112}{7}\\b^4= 16\\b =\sqrt[4]{16}\\b = 2](https://tex.z-dn.net/?f=y%20%3D%20ab%5Ex%5C%5C112%3D7%5Ccdot%20b%5E4%5C%5Cb%5E4%20%3D%20%5Cfrac%7B112%7D%7B7%7D%5C%5Cb%5E4%3D%2016%5C%5Cb%20%3D%5Csqrt%5B4%5D%7B16%7D%5C%5Cb%20%3D%202)
Get the required exponential equation
Recall that
, hence the required equation will be 
Learn more here: brainly.com/question/19245707
Answer:
a. OM is congruent to ON.
Step-by-step explanation:
To use the HL Theorem, you must have a congruent hypotenuse and a congruent leg. In this case, you have two congruent hypotenuses. You just need to find two congruent legs.
a. OM is congruent to ON. This says that two legs are congruent, so this is your answer.
b. LM is congruent to ML. This does not help as it is saying a segment is the same as the same segment.
c. and d. Both of these use angle measurements, which does not help with the HL Theorem.
Hope this helps!
Answer:

Step-by-step explanation:

Cancel the negative signs on both sides.

Evaluate.
