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2 years ago

12

How are number lines and the coordinate plane related?

1 answer:

schepotkina [342]2 years ago

8 0

a coordinate plane is a two-dimensional number line where the vertical line is called the y-axis and the horizontal is called the x-axis. And a number line is
a line on which numbers are marked at intervals, used to illustrate simple numerical operations. So how are the related........?

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What the heck do i do with this

Ad libitum [116K]

Use the law of cosines to find a: $a^2=b^2+c^2-2bc (cos(A))$

So to use the given numbers you get: $a^2=27^2+45^2-2(27)(45)(cos(53))$

Which would come out to equal about 35.94

so a=35.9

7 0

2 years ago

Read 2 more answers

Adam got 97, 70, 89, 99, 100, 78, 79, 89, 77, and 72 on his reading test. What is the mean of Adam's reading test?

ipn [44]

Answer: 85

Add up all the numbers: 97+70+89+99+100+78+79+89+77+72=850

Divide 850 by the amount of numbers:

850/10=85

I hope this helps and good luck! :)

5 0

2 years ago

1. Billy's mother started a college fund on Billy's 8th birthday. She

MaRussiya [10]

When billy turns 18 he will have $3,099.6 in his account because .021 percent x $8,200= $172.20 per year x 18 years = $3,099.60

7 0

2 years ago

The question is in the picture, someone please answer this.

Lemur [1.5K]

Answer:

1) 25x + 15y = 750

2) 25. On a graph it would look like : (25,0)

3) 15. On a graph it would look like : (15,0)

4) 25x - 750 = 15y or 15y - 750 = 25x

Step-by-step explanation:

6 0

2 years ago

Find the area of quadrilateral ABCD. A. 27.28 units² B. 33.08 units² C. 28.53 units² D. 26.47 units²

Ludmilka [50]

Answer:

<h3> $\boxed{ \boxed{ \bold{ \purple{ \sf{28.53 \: {units}^{2} }}}}}$ </h3>

Option C is the correct option.

Step-by-step explanation:

For ABD

$\sf{ = \frac{2.89 + 8.59 + 8.6}{2} }$

$\sf{ = \frac{20.08}{2} }$

$\sf{ = 10.04}$

∆ ABD = $\sf{ = \sqrt{10.04(10.04 - 2.89)(10.04 - 8.59)(10.04 - 8.6)} }$

$\sf{ = \sqrt{10.04 \times 7.15 \times 1.45 \times 1.44 }}$

$\sf{ = \sqrt{149.8891} }$

$\sf{ = 12.2429}$

For ∆ ACD ,

$\sf{s = \frac{8.6 + 4.3 + 7.58}{2} }$

$\sf{ = \frac{20.48}{2} }$

$\sf{ = 10.24}$

∆ ACD = $\sf{ = \sqrt{10.24(10.24 - 8.6)(10.24 - 4.3)(10.24 - 7.58} }$

$\sf{ \sqrt{10.24 \times 1.64 \times 5.94 \times 2.66} }$

$\sf{ = \sqrt{265.3456} }$

$\sf{ = 16.2894}$

Area of quadrilateral ABCD = $\sf{12.2429 + 16.2894}$

$\sf{ = 28.5323}$

$\sf{ = 28.53}$ units ²

Hope I helped!

Best regards!

8 0

3 years ago