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dolphi86 [110]
3 years ago
9

Select all the ordered pairs that are solutions of the inequality

Mathematics
1 answer:
Kaylis [27]3 years ago
8 0

Answer:

The ordered pair solutions are:

C. (1,-3)

E. (2,5)

Step-by-step explanation:

Given inequality:

y

To check from the list of ordered pairs which are the solution of the inequality.

Solution:

In order to check for the solutions to the inequality, we will plugin the ordered pairs in the inequality and check if it holds true. Each ordered pair is in the form (x,y).

Plugging in each ordered pair in the given inequality to find the solutions.

<em>A) (-3,-2)</em>

-2

-2

-2

The above statement can never be true and hence it is not a solution.

<em>B) (3,14)</em>

14

14

14

The above statement can never be true and hence it is not a solution.

<em>C) (1,-3)</em>

-3

-3

-3

<u>The above statement is always true and hence it is a solution.</u>

<em>D) (1,6)</em>

6

6

6

The above statement can never be true and hence it is not a solution.

<em>C) (2,5)</em>

5

5

5

<u>The above statement is always true and hence it is a solution.</u>

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The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am
posledela

Step-by-step walkthrough:

a.

Well a standard half-life equation looks like this.

N = N_0 * (\frac{1}{2})^{t/p

N_0 is the starting amount of parent element.

N is the end amount of parent element

t is the time elapsed

p is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

N(t) = 74g * (\frac{1}{2})^{t/2.8days

b.

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2}  =  37g

c.

Now we just gotta do some algebra. Use the original function but this time, replace N(t) with 10g and solve algebraically.

10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}

Take the log of both sides.

log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})

Use the exponent rule for log laws that, log(b^x) = x*log(b)

log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})

\frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = \frac{t}{2.8days}

2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = t

slap that in your calculator and you get

t = 8.1 days

7 0
2 years ago
Three-fifths (3/5) of a foot of rain was collected in two-thirds (2/3) of a day. At that rate, how much rain would be collected
pickupchik [31]
Nine-tenths (9/10) of a foot of rain would be collected in one day.
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3 years ago
What vaule of x makes th question true? 3x+2(x-5)=50 a. 8 b.9 c.11 d.12
Ainat [17]
Hey Tierra7l, 

Lets solve your equation together step by step. 

3x+2\left(x-5\right)=50 

\mathrm{Expand}\:2\left(x-5\right):\quad 2x-10 

\mathrm{Distribute\:parentheses\:using}:\quad \:a\left(b+c\right)=ab+ac 

                                    a=2,\:b=x,\:c=-5 

2\cdot \:x+2\left(-5\right) 

\mathrm{Apply\:minus-plus\:rules} 

\:\:+\left(-a\right)=-a 

2x-2\cdot \:5 

\mathrm{Multiply\:the\:numbers:}\:2\cdot \:5=10 

2x-10 

3x+2x-10=50 

\mathrm{Add\:similar\:elements:}\:3x+2x=5x 

5x-10=50 

\mathrm{Add\:}10\mathrm{\:to\:both\:sides} 

5x-10+10=50+10 

\mathrm{Simplify} 

5x=60 

\mathrm{Divide\:both\:sides\:by\:}5 

\dfrac{5x}{5}=\dfrac{60}{5} 

\mathrm{Simplify} 

x=12 

Hope this helps, 

      AnthrαX <span>  </span>
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Show me how to solve -6+x/4= -5
schepotkina [342]
Multiply 4(-5) then add 6 and theres your X

7 0
3 years ago
Cara is a high school student. On average, there are 26 students per class at her school, with a standard deviation of 3 student
aliina [53]

Answer:

B. 97%

Step-by-step explanation:

20 students is -2 standard deviations away from the mean

35 is +3 standard deviations away from the mean

100 - 2.4 = 97.6

(2.4 is amount of data we are not using)

please refer to the standard deviation graph attached :)

8 0
2 years ago
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