For this case we have that by definition:

So:

We apply distributive property that states that:

In addition, we take into account that:


Answer:

Option B
Answer:
C=hypotenuse [h]=50 feet
A=[base]=40foot.
B=perpendicular [p]=?
Area =?
we have
By using Pythagoras law
p²+b²=h²
B²=40²=50²
B²=50²-40²
B=√900
B=30feet
we have
Area of triangle =½×A×B=½×30×40=600feet²
<u>the roped off </u><u>area</u><u> </u><u>is</u><u> </u><u>6</u><u>0</u><u>0</u><u> </u><u>square</u><u> </u><u>feet</u><u>.</u>
<u>it</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>right</u><u> </u><u>angled</u><u> </u><u>triangle</u><u>.</u>
Answer:
130
Step-by-step explanation:
From the FIQURE of the question, we can see that line "e" is parallel to line "f" then we can say (2x + 18)=( 4x - 14)
Then we can collect like terms
2x-4x= -14-18
-2x=-32
The negatives cancelled out we have
x = 16
From the question, At the intersection of lines b and f, the top right angle is (4 x minus 14) degrees
Angle on a straight line is 180° then we can calculate "y" as
y = 180 - (4x - 14)
= 180-4x+14
= 180+14-4x
But x= 16, then substitute we have
= 180 +14 - 14(16)
= 130
Hence, the value of y is 130
CHECK THE ATTACHMENT FOR THE FIQURE
Answer:
10.50°C
Step-by-step explanation:
Given x = 2 + t , y = 1 + 1/2t where x and y are measured in centimeters. Also, the temperature function satisfies Tx(2, 2) = 9 and Ty(2, 2) = 3
The rate of change in temperature of the bug path can be expressed using the composite formula:
dT/dt = Tx(dx/dt) + Ty(dy/dt)
If x = 2+t; dx/dt = 1
If y = 1+12t; dy/dt = 1/2
Substituting the parameters gotten into dT/dt we will have;
dT/dt = 9(1)+3(1/2)
dT/dt = 9+1.5
dT/dt = 10.50°C/s
Hence the rate at which the temperature is rising along the bug's path is 10.50°C/s