Answer:
NUMBER 1.)
Step 1
Subtract 3y3y from both sides.
5x=10-3y5x=10−3y
Step 2
Divide both sides by 55.
\frac{5x}{5}=\frac{10-3y}{5}
5
5x
=
5
10−3y
Hint
Undo multiplication by dividing both sides by one factor.
Step 3
Dividing by 55 undoes the multiplication by 55.
x=\frac{10-3y}{5}x=
5
10−3y
Hint
Undo multiplication.
Step 4
Divide 10-3y10−3y by 55.
x=-\frac{3y}{5}+2x=−
5
3y
+2
Hint
Divide.
Solution
x=-\frac{3y}{5}+2x=−5
3y+2
Step-by-step explanation:
NUMBER 2.)
Step 1
Add 4y4y to both sides.
3x=6+4y3x=6+4y
Step 2
The equation is in standard form.
3x=4y+63x=4y+6
Step 3
Divide both sides by 33.
\frac{3x}{3}=\frac{4y+6}{3}
3
3x
=
3
4y+6
Hint
Undo multiplication by dividing both sides by one factor.
Step 4
Dividing by 33 undoes the multiplication by 33.
x=\frac{4y+6}{3}x=
3
4y+6
Hint
Undo multiplication.
Step 5
Divide 6+4y6+4y by 33.
x=\frac{4y}{3}+2x=
3
4y
+2
Hint
Divide.
Solution
x=\frac{4y}{3}+2x= 3
4y+2
Answer:
C)28in
Step-by-step explanation:
To get the area of this face, we will divide it into 3 sections.
Area of the 1st section:
Given:
l=4in
w=2in
Formula:
a=l*w
Solution:
a=l*w
a=8in^2
Area of the 2nd section:
Given:
l=2in
w=2in
Formula:
a=l*w
Solution:
a=l*w
a=2in*2in
a=4in^2
Area of the 3rd section:
l=8in
w=2in
Formula:
a=l*w
Solution:
a=l*w
a=8in*2in
a=16in^2
a1+a2+a3=Area of the face
8in^2+4in^2+16in^2=28in^2
Hope this helps ;) ❤❤❤
<span>The area in the right tail more extreme than z= 3.01 in a standard normal distribution is given by
P(z > 3.01) = 1 - P(z < 3.01) = 1 - 0.99869 = 0.00131
</span>
Sum of all exterior angle should be 360.
when the measure of each exterior angle is 36 then it has 10 sides and it is a
<span>Decagon </span>
Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.
We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)
N = -(sin7A - sinA) + sin5A - sin3A
= -2cos4A*sin3A + 2cos4A*sinA
= 2cos4A(sinA - sin3A)
= 2cos4A*2cos(2A)sin(-A)
= -4cos4A*cos2A*sinA
D = cos7A + cosA - (cos5A + cos3A)
= 2cos4A*cos3A - 2cos4A*cosA
= 2cos4A(cos3A - cosA)
= 2cos4A*(-2)sin2A*sinA
= -4cos4A*sin2A*sinA
Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
= cos2A/sin2A
= cot2A
This verifies the identity.
