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2 years ago

Which ordered pairs make the open sentence true?

Mathematics

1 answer:

My name is Ann [436]2 years ago

4 0

The ordered pairs which make the open sentence true are:

(-1, 6).
(4, -1).

<h3>What is an inequality?</h3>

An inequality can be defined as a mathematical relation that compares two (2) or more integers and variables in an algebraic expression based on any of the following arguments:

Less than (<).
Greater than (>).
Less than or equal to (≤).
Greater than or equal to (≥).

For this exercise, you should evaluate each of the given expressions to determine which inequalities is true when the ordered pairs are substituted. This ultimately implies that, you would have to substitute the ordered pairs into each of the given algebraic expressions and then evaluate.

When ordered pairs = (3, 6), we have:

3x + y < 14

3(3) + 6 < 14

9 + 6 < 14

15 < 14 (False).

When ordered pairs = (7, -7), we have:

3x + y < 14

3(7) + (-7) < 14

21 - 7 < 14

14 < 14 (False).

When ordered pairs = (6, 0), we have:

3x + y < 14

3(6) + 0 < 14

18 - 0 < 14

18 < 14 (False).

When ordered pairs = (-1, 6), we have:

3x + y < 14

3(-1) + 6 < 14

-3 + 6 < 14

3 < 14 (True).

When ordered pairs = (4, -1), we have:

3x + y < 14

3(4) + (-1) < 14

12 - 1 < 14

11 < 14 (True).

Read more on ordered pairs here: brainly.com/question/22683073

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1.048576e+46

Step-by-step explanation:

(5x5)x20x80 to the power of 10

(25x20)x80 to the power of 10

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40000 to the power of 10

the answer is 1.048576e+46

8 0

3 years ago

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Anastaziya [24]

Answer:

UHHH its 1/2

Step-by-step explanation:

because it goes up 1 and run 2.

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3 years ago

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2) 83,97,85,84,96, 80,80,87,91<br> Mean median and mode

marusya05 [52]

$\text {Hi! Let's Solve this Problem!}$

$\text {Before you find the Mean, Medina and Mode you must put the numbers in order.}$

$\text {Before: 83, 97, 85, 84, 96, 80, 80, 87, 91}\\\text {After: 80, 80, 83, 84, 85, 87, 91, 96, 97}$

$\underline {\text {Mean}}$

$\text {To Find the Mean you must Add all the Numbers then Divide.}$

$\text {Add:}$ $\text 80+80+83+84+85+87+91+96+97=\fbox {783}}$

$\text {Divide:}$ $\text {783/9=}$

$\text {The Mean Is:}$

$\fbox {87}$

$\underline {\text Median}}$

$\text {To find the Median you find the number that's in the middle of the set.}$

$\text {Put your Left Pointer Finger on 83. Put your Right Pointer Finger on 91.}$

$\text {Move your Left Finger to the Right. Move your Right Finger to the Left.}$

$\text {Once your Left Finger makes it to 84 and your Right Finger makes it to 87}$ $\text {it shows that 85 is left in the middle.}$

$\text {The Median Is:}$

$\fbox {85}$

$\underline {\text {Mode}}$

$\text {Finding the Mode means finding the number}$ $\text { that shows the most in the number set given.}$

$\text {Since 80 is the number that is showing more than any other number}$ $\text {this tells us that 80 is our Mode.}$

$\text {The Mode Is:}$

$\fbox {80}$

$\text {Best of Luck!}$

4 0

4 years ago

the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?

madam [21]

Answer:

About 0.6548 grams will be remaining.

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

$f(t)=a(r)^t$

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

$\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}$

One year has 365 days.

Therefore, the amount remaining after one year will be:

$\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548$

About 0.6548 grams will be remaining.

Alternatively, we can use the standard exponential growth/decay function modeled by:

$f(t)=Ce^{kt}$

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

$255=510e^{38k}$

Solving for k:

$\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)$

Thus, our function is:

$f(t)=510e^{t\ln(.5)/38}$

Then after one year or 365 days, the amount remaining will be about:

$f(365)=510e^{365\ln(.5)/38}\approx 0.6548$

5 0

3 years ago

Write an equation in standard firm using integers y-4=5(x-8)

diamong [38]

I would say its y=5x-36

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3 years ago