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3 years ago

Is 3(x-2) equivalent to 3x-6

Mathematics

1 answer:

Shtirlitz [24]3 years ago

3 0

Answer:

yes that would be equivalent to 3x-6

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Kyle entered to run in a 5K (5 kilometers) race. He wants to figure out how many miles are in the race. If there are 1000 meters

tekilochka [14]

Answer:

3.11 miles

Step-by-step explanation:

Hello!

First we have to get the distance into meters

The problem says there are 5 kilometers in a race and 1000 meters in a kilometer so 5 kilometers is 5000 meters

Now we turn the meters into miles

It says a mile is about 1609.34 so we can divide the amount of meters by how many meters are in a mile

5000/1609.34 = 3.106

Round to the nearest tenth

3.106 ⇒ 3.11

The answer is 3.11 miles

Hope this helps!

8 0

4 years ago

Noah took pictures of different lizards for a science project. He wanted 20 pictures altogether, but he only had 11 pictures so

NNADVOKAT [17]

Answer:

9 pictures

Step-by-step explanation:

if Noah needed 20 pictures in all ,but only had 11 he would subtract 11 from 20 leaving Noah with 9 pictures.

6 0

3 years ago

Draw the graph of each line y=2x-3

torisob [31]

Step-by-step explanation:

Here is the answer for your question

5 0

3 years ago

Read 2 more answers

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

4 years ago

Price of automobile= $3,000

AysviL [449]

About 36000 because if u multiply 3000 the price of the automobile by number of payments 12 you get 36000

7 0

3 years ago