Answer:
x=-4
Step-by-step explanation:
(x+4) ^ (1/3) + (2x+8) ^ (1/3) = 0
Subtract (x+4) ^ (1/3) from each side
(x+4) ^ (1/3) - (x+4) ^ (1/3)+ (2x+8) ^ (1/3) = -(x+4) ^ (1/3)
(2x+8) ^ (1/3) = -(x+4) ^ (1/3)
Cube each side
(2x+8) ^ (1/3) ^3= -(x+4) ^ (1/3)^3
2x+8 = -(x+4)
Distribute the negative sign
2x+8 = -x -4
Add x to each side
2x+8 +x =-x+x-4
3x+8 = -4
Subtract 8 from each side
3x+8-8 =-4-8
3x =-12
Divide by 3
3x/3 = -12/3
x = -4
Answer:
In Δ CFD , CD is the LONGEST side.
Step-by-step explanation:
Here, the given Δ CSD is a RIGHT ANGLED TRIANGLE.
Now, as we know in a right triangle, HYPOTENUSE IS THE LONGEST SIDE.
So, in Δ CSD SD is the longest side as SD = Hypotenuse.
Now, an altitude CF is drawn to hypotenuse SD.
⇒ CF ⊥ SD
⇒ Δ CFD is a RIGHT ANGLED TRIANGLE with ∠ F = 90°
and CD as a hypotenuse.
⇒ In Δ CFD , CD is the LONGEST side.
Hence, CD is the longest side in the given triangle CFD.
Equation of the given line will be :
now, let's plug the value of x as the x - coordinate of the above point (-2 , 9) to check whether it lies on the line or not.
now, since value of isn't equal to the y - coordinate of the given point, so the point doesn't lies on the line.
Answer:
Part A) Option A. QR= 3 cm
Part B) Option B. SV=6.5 cm
Step-by-step explanation:
step 1
<u>Find the length of segment QR</u>
we know that
If two triangles are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent
so
In this problem Triangle QRW and Triangle QSV are similar by AA Similarity Theorem
so

we have
---> because S is the midpoint QT (QS=TS)
--->because V is the midpoint QU (QW+WV=VU)
--->because V is the midpoint QU (QV=VU)
substitute the given values

solve for QR

step 2
Find the length side SV
we know that
The <u><em>Mid-segment Theorem</em></u> states that the mid-segment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle, and the length of this mid-segment is half the length of the third side
so
In this problem
S is the mid-point side QT and V is the mid-point side QU
therefore
SV is parallel to TU
and

so
