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4 years ago

12

Does someone know where on the quadrant go on the coordinate planes

1 answer:

stiv31 [10]4 years ago

6 0

The origin is at 0 on the x-axis and 0 on the y-axis. The intersecting x- and y-axes divide the coordinate plane into four sections. These four sections are called quadrants. Quadrants are named using the Roman numerals I, II, III, and IV beginning with the top right quadrant and moving counter clockwise.

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Choose a value for n, and then evaluate n^2.

Sloan [31]

Pick your favorite number and square it.

Squaring a number means you multiply it by itself.

Let's say you picked the number 7

7 squared = 7^2 = 7*7 = 49

Think of a 7 by 7 grid of tiny squares. There would be 49 total little squares to represent the total area. Graph paper might help visualize this.

6 0

2 years ago

Liliana and Marcus are playing a word game. Liliana spells a word incorrectly and loses 18 points. Then Marcus spells a word cor

Pepsi [2]

Answer:

Well if Liliana loses 18 points and Marcus get the correct one and gains 18 points

Step-by-step explanation:

So basically it's going to be negative because its saying what is the opposite of Liliana's core but in reverse order

3 0

2 years ago

Given independent events A and B such that P(A) = 0.3 and P (B) = 0.5, which of the following is a correct statement?

aev [14]

Answer:

(D) P(AUB) = 0.65

Step-by-step explanation:

Since A and B are independent:

P(A ^ B)

= P(A) × P(B)

= 0.3 × 0.5 = 0.15

P(AUB) = P(A) + P(B) - P(A^B)

= 0.3 + 0.5 - 0.15

= 0.65

7 0

3 years ago

What is the answer for pie in math

salantis [7]

Answer:

first how do I change my name

5 0

3 years ago

Read 2 more answers

Use the following image to assist in your answer.<br> a =<br> - 6<br> - 9<br> - 4

vfiekz [6]

Step-by-step explanation:

Pythagoras' theorem for the smallest one :

$c^2 = 4^2 + 6^2$

$c^2 = 16 + 36$

$c^{2}$ = 52

Pythagoras' theorem for the middle one :

$b^{2}$ = $6^{2}$ + $a^{2}$

Pythagoras' theorem for the biggest one :

$(4+a)^2 = c^2 + b^2$

$16 + 8a + a^2 = 52 + b^2$

Using the formula before (for $b^2$ ) it becomes :

$16 + 8a + a^2 = 52 + (6^2 + a^2)$

$16 + 8a + a^2 = 52 + 6^2 + a^2$

$16 + 8a = 52 + 6^2$

16 + 8a = 52 + 36

16+8a = 88

8a = 88-16

8a = 72

$a = \frac{72}{8}$

a = 9

Verifying :

$b^2 = 6^2 + a^2$

$b^2 = 36 + 81$

$b^2 = 117$

$b^{2}$ = 117

The biggest one :

$(4+a)^2 = c^2 + b^2$

$(4+9)^2 = 52 + 117$

$13^2 = 169$

True

8 0

3 years ago