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Debora [2.8K]
3 years ago
14

Help! I need this for tomorrow!

Mathematics
1 answer:
Masja [62]3 years ago
4 0
1. 1/2- you count the total (8) and then count the two and all the numbers greater than 5( don’t count 5) that is 4 so it would be 4/8 simplified to 1/2

2. 1/2- count the total (still 8) and how many prime numbers there are (4- 2,3,5, and 7) the answer is 4/8 simplified to 1/2
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What is the distance between P(-3,-25) and Q(50,-2)
Karolina [17]
Dist =   (x2-x1)^{2} +  (y2-y1)^{2}

above  is the distance formula, plugging your numbers from P(x1,y1) and Q(x2,y2).

Dist =  \sqrt{(50-(-3))^{2}  + (-2-(-25))^{2} }
or\sqrt{ 53^{2} + 23 ^{2} }


4 0
2 years ago
What is the independent and dependent variable in this situation?
lord [1]
Cid Sam is a way to remember
C -control S - SAME (keep the same)

I -independent A- alter

D-depend M- measure

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8 0
3 years ago
Please help this is so confusing u can have 25 points if u help me
Elina [12.6K]

Step-by-step explanation:

Share 20pounds in the ratio 2:3

Total ratio=2+3=5

2/5×20=8

20-8=12

Sharing 20pounds in that ratio gives 8pounds:12pounds

Share 15cm in the ratio 1:3

Total ratio=1+4=4

¼×15=3.75

15-3.75=11.25

Sharing 15cm in that ratio gives 3.75:11.25

7 0
2 years ago
Distributive property (-2)(a 6)
aivan3 [116]
Hello there.

<span>Distributive property (-2)(a 6)

-2a^6</span>
5 0
3 years ago
Solving Exponential and Logarithmic Equations In Exercise, solve for x.<br> 500(1.075)120x = 100.000
Volgvan

Answer:

The solution is:

x = 0.61

Step-by-step explanation:

The first step to solve this equation is placing everything with the exponential to one side of the equality, and everything without the exponential to the other side. So

500(1.075)^{120x} = 100000

(1.075)^{120x} = \frac{100000}{500}

(1.075)^{120x} = 200

To find x, we have to apply log to both sides of the equality.

We also have that:

\log{a^{x}} = x\log{a}

So

\log{(1.075)^{120x}} = \log{200}

120x\log{1.075} = 2.30

120x*0.03 = 2.30

3.77x = 2.30

x = \frac{2.30}{3.77}

x = 0.61

4 0
3 years ago
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