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3 years ago

Which of the following functions has the function rule y = x + 4?

Mathematics

2 answers:

hichkok12 [17]3 years ago

8 0

The correct answer is A

Nataly_w [17]3 years ago

8 0

The correct answer is “a”

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Indicate in standard form the equation of the line through the given points. K(6, 4), L(-6, 4)

azamat

y=4

the values of x changes while the y doesn't so it's y=4

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4 years ago

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What is the slope of a line that is parallel to the x-axis? 0 1 undefined

Troyanec [42]

Its 0 because the slope would be one if it was a positive line and itd go up and across one unitl and it is undefined if it is parallel to the y axis

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3 years ago

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Graph (x+2)^2+y^2=100

LenaWriter [7]

Answer:

circle of radius 10 centered at (-2, 0)

Step-by-step explanation:

The equation is of the form ...

(x -h)^2 +(y -k)^2 = r^2

with (h, k) = (-2, 0) and r=10.

This is the standard form of the equation of a circle with radius r centered at (h, k).

See the attached for a graph.

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4 years ago

Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (n(2n-

Charra [1.4K]

Answer:

The statement is true is for any $n\in \mathbb{N}$ .

Step-by-step explanation:

First, we check the identity for $n = 1$ :

$(2\cdot 1 - 1)^{2} = \frac{2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1)}{3}$

$1 = \frac{1\cdot 1\cdot 3}{3}$

$1 = 1$

The statement is true for $n = 1$ .

Then, we have to check that identity is true for $n = k+1$ , under the assumption that $n = k$ is true:

$(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot (2\cdot k +1)^{2} = (k+1)\cdot (2\cdot k +1)\cdot (2\cdot k +3)$

$(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)$

$k\cdot (2\cdot k - 1)+3\cdot (2\cdot k +1) = (k + 1)\cdot (2\cdot k +3)$

$2\cdot k^{2}+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)$

$(k+1)\cdot (2\cdot k + 3) = (k+1)\cdot (2\cdot k + 3)$

Therefore, the statement is true for any $n\in \mathbb{N}$ .

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3 years ago

David is the coach of the Bingtown Cougars hockey team. There are 2 minutes left in the game they are playing and they are curre

sweet [91]

Answer:

Step-by-step explanation:

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3 years ago