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3 years ago

Which point is the same distance from the y-axis but on the opposite side of the y-axis as point C?

Mathematics

2 answers:

Margaret [11]3 years ago

8 0

Answer:

5.2 i think

Step-by-step explanation:

bixtya [17]3 years ago

5 0

Answer: The answer would be 5.2

Step-by-step explanation:

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Determine whether each set of side lengths could be the sides of a right triangle. Drag and drop each set of side lengths to the

Neko [114]

Answer:

10.5cm,20.8cm,23.3cm — yes
6cm, 22.9cm,20.1cm — no

Step-by-step explanation:

If the sides form a right triangle, the sum of the squares of the shorter two sides will equal the square of the longest side.

1. 10.5^2 + 20.8^2 = 23.3^2 . . . . . true algebraic statement; right triangle

2. 6^2 +20.1^2 = 440.01 ≠ 22.9^2 = 524.41 . . . . . this is an obtuse triangle

7 0

3 years ago

What is the value of the fourth term in the geometric sequence

ivanzaharov [21]

Answer:

a(4) = 15/4

Step-by-step explanation:

Here we're told that the first term is a(1) = 30 and that the common factor r = 1/2.

Thus, the geometric sequence formula specific to this case is

a(n) = 30(1/:2)^(n-1).

What is the fourth term? Let n = 4,

a(4) = 30(1/2)^(4-1), or a(4) = 30(1/2)^(3), or a(4) = 30(1/8) = 30/8, or, in reduced form,

a(4) = 15/4.

6 0

3 years ago

Show both decimal places (5.06).

lilavasa [31]

a. the cost of a call for six minutes is 0.70.

b.the cost of a 14-minute call is 2.62.

c.the cost of a 9 ½2-minute call is 1.66.

given that

rate schedule for an m-minute call from any of its pay phones 0.70.

c(m) = 0.70 when m<=6

c(m) = 0.70+ 0.24(m-6) when m>6 & m is an integer

c(m) = 0.70+0.24((m-6)+1) when m>6 & m is not an integer.

a. to find the cost of a call for six minutes.

already given that c(m) =0.70 when m=6.

Therefore, the cost of a call for six minutes is 0.70.

b. to find the cost of a 14-minute call.

according to given information

$c(14) = 0.70 + 0.24(m - 6) \\ c(14) = 0.70 + 0.24(14 \: - 6) \\c(14) = 0.70 + (0.24 \times 14) - (0.24 \times 6) \\c(14) = 0.70 + (3.36 - 1.44) \\ c(14) = 0.70 + 1.92 \\ c(14) = 2.62$

Therefore,the cost of a 14-minute call is 2.62.

c.to find the cost of a 9 ½2-minute call

according to given information

$c(9 \frac{1}{2} 2) = 0.70 + 0.24((9 \frac{1}{2} 2 - 6) + 1) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.24((9 - 6) + 1) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.24(3+ 1) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.24(4) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.96 \\ c(9 \frac{1}{2} 2) = 1.66$