Answer:
d. 6
Step-by-step explanation:
The product of lengths to the near and far intersection points is the same for both secants:
PB×PA = PD×PC
We can use this relation to solve for x, then use the value of x to find PD.
7×12 = (3x)(7x)
4 = x² . . . . . . . . . divide by 21
2 = x . . . . . . . . . . take the square root
__
PD = 3x = 3·2
PD = 6
Answer:
The answer is 1.79 × 10-3
Answer: 14
Step-by-step explanation:
Input Data :
Point 1 (
x
A
,
y
A
) = (-4, -8)
Point 2 (
x
B
,
y
B
) = (10, -8)
Objective :
Find the distance between two given points on a line?
Formula :
Distance between two points = √
(x
B
−
x
A
)
2
+
(
y
B
−
y
A
)
2
Solution :
Distance between two points = √
(
10
− −
4
)
2
+
(
−
8
− −
8
)
2
= √
14
^2
+
0
^2
= √
196
+
0
= √
196
= 14
Distance between points (-4, -8) and (10, -8) is 14
The formula for the nth term of a geometric sequence:
a₁ - the first term, r - the common ratio
![54, a_2, a_3, 128 \\ \\ a_1=54 \\ a_4=128 \\ \\ a_n=a_1 \times r^{n-1} \\ a_4=a_1 \times r^3 \\ 128=54 \times r^3 \\ \frac{128}{54}=r^3 \\ \frac{128 \div 2}{54 \div 2}=r^3 \\ \frac{64}{27}=r^3 \\ \sqrt[3]{\frac{64}{27}}=\sqrt[3]{r^3} \\ \frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\ r=\frac{4}{3}](https://tex.z-dn.net/?f=54%2C%20a_2%2C%20a_3%2C%20128%20%5C%5C%20%5C%5C%0Aa_1%3D54%20%5C%5C%0Aa_4%3D128%20%5C%5C%20%5C%5C%0Aa_n%3Da_1%20%5Ctimes%20r%5E%7Bn-1%7D%20%5C%5C%0Aa_4%3Da_1%20%5Ctimes%20r%5E3%20%5C%5C%0A128%3D54%20%5Ctimes%20r%5E3%20%5C%5C%0A%5Cfrac%7B128%7D%7B54%7D%3Dr%5E3%20%5C%5C%20%5Cfrac%7B128%20%5Cdiv%202%7D%7B54%20%5Cdiv%202%7D%3Dr%5E3%20%5C%5C%0A%5Cfrac%7B64%7D%7B27%7D%3Dr%5E3%20%5C%5C%0A%5Csqrt%5B3%5D%7B%5Cfrac%7B64%7D%7B27%7D%7D%3D%5Csqrt%5B3%5D%7Br%5E3%7D%20%5C%5C%0A%5Cfrac%7B%5Csqrt%5B3%5D%7B64%7D%7D%7B%5Csqrt%5B3%5D%7B27%7D%7D%3Dr%20%5C%5C%0Ar%3D%5Cfrac%7B4%7D%7B3%7D)
