In the case above, Talib work is not correct as one need to first switch x and y before one can solve for y.
<h3>What is the variables about?</h3>
Note that:
y=-8x+4
y-4=-8x
(y-4)/-8=x
Since the independent variable x is known, one can switch the variable labels and thus it will be:
y=(x-4)/-8
f^-1(x)=(x-4)/-8
This can be written again as:
f^-1(x)=(4-x)/8 :P
Thus one can say No, as he forgot to switch the variable labels after solving for the independent variable.
In the case above, Talib work is not correct as one need to first switch x and y before one can solve for y.
See the first part of the question below
Talib is trying to find the inverse of the function to the right. His work appears beneath it. Is his work correct? Explain your answer.
Learn more about variable from
brainly.com/question/4063302
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Answer:
Step-by-step explanation:
Using the section formula, if a point (x,y) divides the line joining the points (x
1
,y
1
) and (x
2
,y
2
) in the ratio m:n, then
(x,y)=(
m+n
mx
2
+nx
1
,
m+n
my
2
+ny
1
)
The vertices of the triangle are given to be (x
1
,y
1
),(x
2
,y
2
) and (x
3
,y
3
). Let these vertices be A,B and C respectively.
Then the coordinates of the point P that divides AB in l:k will be
(
l+k
lx
2
+kx
1
,
l+k
ly
2
+ky
1
)
The coordinates of point which divides PC in m:k+l will be
⎩
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎧
m+k+l
mx
3
+(k+l)
(l+k)
lx
2
+kx
1
,
m+k+l
my
3
+(k+l)
(l+k)
ly
2
+ky
1
⎭
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎫
⇒(
m+k+l
kx
1
+lx
2
+mx
3
,
m+k+l
ky
1
+ly
2
+my
3
)
The original area of a face would be a^2. Now that you added b to the edge, the new area of each face would be (a+b)^2. To find how much the are increased, subtract a^2 from (a+b)^2. 
So the answer is b(2a+b)
Answer:
7^9
Step-by-step explanation:
<u>Explanation</u><u>:</u>
Consider ABCD is a rhombus
We know that
All sides are equal in rhombus i.e,
⇛AB=BC=CD=DA
and AC and BD are digonals
Given that
Diagonal and the side of the rhombus are equal.
⇛AB = BC = CD = DA = AC
Diagonal AC divides the rhombus into two triangles .
They are ∆ BAC and ∆ DAC
In triangle BAC
BA=BC=AC,(Given)
⇛∠ BAC=∠ABC= ∠ACB =60°→→→Eqn(i)
Similarly in ∆DAC ,
DA=DC=AC
⇛∠DAC=∠ACD=∠ADC=60°→→→Eqn(ii)
From eqn(i) and eqn(ii)
∠A=∠BAC+∠DAC=60°+60°=120°
and
∠B= ∠ABC = 60°.
and
∠C=∠ACB+∠ACD=60°+60°=120°
and
∠D =∠ADC=60°
∴ ∠A = 120° , ∠B = 60° ,∠C = 120° & ∠D = 60°
<u>Answer:</u><u>-</u>The measures of the all angles in the rhombus are 120° , 60° ,120° and 60°.
Note: [Figure refers in the attached file.
