Answer:
B
Step-by-step explanation:
![6^{\frac{1}{4} } b^{\frac{3}{4} }c^{\frac{1}{4} }\\\\=(6^1b^3c^1)^{\frac{1}{4} }\\\\=(6b^3c)^\frac{1}{4} \\\\=\sqrt[4]{6b^3c}](https://tex.z-dn.net/?f=6%5E%7B%5Cfrac%7B1%7D%7B4%7D%20%7D%20b%5E%7B%5Cfrac%7B3%7D%7B4%7D%20%7Dc%5E%7B%5Cfrac%7B1%7D%7B4%7D%20%7D%5C%5C%5C%5C%3D%286%5E1b%5E3c%5E1%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%20%7D%5C%5C%5C%5C%3D%286b%5E3c%29%5E%5Cfrac%7B1%7D%7B4%7D%20%5C%5C%5C%5C%3D%5Csqrt%5B4%5D%7B6b%5E3c%7D)
so answer is B
Answer:
d.
Step-by-step explanation:
if x varies directly as y,
x = ky
8 = 12k
12k = 8
k = 8/ 12
k = 2/3,
x = 2/3 y
now substitute the value of any of the numbers of each pair, and check if it works:
for e.g, find x when y = 15
x = 2/3y
x = 2/3 × 15
x = 10
since you obtain x = 10 when y = 15, the pair is a possible corresponding value of x and y. now do the same for the rest .
ii) x = 2/3y
x = 2/3 × 3, when y = 3
x = 2
hence, this is also a possible pair of x and y values.
iii) x = 2/3y
x = 2/3 × 9 when y = 9
x = 6
again, a possible pair of x and y values.
iv) x = 2/3y
x = 2/3 × 20 when y = 20
x = 40/3
here, when y = 20, x is not 15. it is 40/3.
thus, it can be determined that this is not a possible pair of x and y values.
please try to understand and have a great day!
G(9,-3) being dilated by a scale factor of 1/3 is G'(3,-1) so D. All I did was divide the coordinates by 3 and there you go.
Answer:
153.9
Step-by-step explanation:
Answer:
Given statement is TRUE.
Step-by-step explanation:
Given that line segment JK and LM are parallel. From picture we see that LK is transversal line.
We know that corresponding angles formed by transversal line are congruent.
Hence ∠JKL = ∠ MLK ...(i)
Now consider triangles JKL and MLK
JK = LM {Given}
∠JKL = ∠ MLK { Using (i) }
KL = KL {common sides}
Hence by SAS property of congruency of triangles, ΔJKL and ΔMLK are congruent.
Hence given statement is TRUE.
