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3 years ago

Determine whether the system has no one, or infinitely many solutions.

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

6 0

This has no solution

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Chuy wants to buy a new television. The television costs $1,350. Chuy decides to save the same amount of money each week, for 27

Korvikt [17]

Answer: C: Chuy will save more than he needs and will meet his goal at the end of 27 weeks.

Step-by-step explanation: Divide the amount of money saved, $440, by the number of weeks, 8.

440/8 = $55.

This means $55 is the amount of money Chuy saves every week.

Multiply $55 by the number of weeks Chuy plans to save money, 27.

55 x 27 = 1,485.

Since the television costs $1,350 dollars, Chuy will save more than he needs and will meet his goal in less than 27 weeks.

4 0

3 years ago

When a fraction of 12 is taken away from 17, what remain exceed one-third of seventeen by six.

vova2212 [387]

X - the fraction

$17- 12x=\frac{1}{3} \times 17+6 \\
17-12x=\frac{17}{3}+6 \\
-12x=\frac{17}{3}+6-17 \\
-12x=\frac{17}{3}-11 \\
-12x=\frac{17}{3}-\frac{33}{3} \\
-12x=-\frac{16}{3} \\
x=-\frac{16}{3} \times (-\frac{1}{12}) \\
x=\frac{16}{36} \\
x=\frac{4}{9}$

The fraction is 4/9.

5 0

3 years ago

How to find the area of a polygon when u dont know the side length or tge apothem

dlinn [17]

Measure it hope that helps if not that's on me

3 0

3 years ago

For the person who were helping me

Fittoniya [83]

Take 1600 and 85% (turn 85% into a decimal .85) and multiply them together. you should get that 1360 tickets were sold. hope this helps and please give me the brainliest answer.

5 0

3 years ago

A tank initially contains 40 ounces of salt mixed in 100 gallons of water. a solution containing 4 oz of salt per gallon is then

stellarik [79]

Let $A(t)$ be the amount of salt in the tank at time $t$ . We're given that $A(0)=40$ . The rate at which this amount changes is given by

$A'(t)=\dfrac{5\text{ gal}}{1\text{ min}}\cdot\dfrac{4\text{ oz}}{1\text{ gal}}-\dfrac{5\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ oz}}{100+(5-5)t\text{ gal}}$

$A'(t)+\dfrac{A(t)}{20}=20$

$e^{t/20}A'(t)+e^{t/20}\dfrac{A(t)}{20}=20e^{t/20}$

$\bigg(e^{t/20}A(t)\bigg)'=20e^{t/20}$

$e^{t/20}A(t)=400e^{t/20}+C$

$A(t)=400+Ce^{-t/20}$

Since $A(0)=40$ , we get

$40=400+C\implies C=-360$

so that the amount of salt at time $t$ is

$A(t)=400-360e^{-t/20}$

After 20 minutes, the tank contains

$A(20)=400-360e^{-20/20}\approx267.56\text{ oz}$

6 0

3 years ago