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3 years ago

What is the solution to this equation?

Mathematics

1 answer:

slamgirl [31]3 years ago

4 0

Answer: No solution.

Step-by-step explanation:

1. The lines are parallel.

2. If you solve the system of equations you would end up with 0x, so you cannot solve for x.

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What doexs this equal: 3ab-9ab+7ab

AfilCa [17]

Answer:

I think that it equals 1ab. Because 3ab - 9ab= -6ab + 7ab= 1ab

8 0

3 years ago

Read 2 more answers

Please help ASAP. Need it for homework

kondaur [170]

Answer:

Speedy Taxis

Step-by-step explanation:

I solved this by using a graphing calculator. Speedy Taxis shows the least cost for x=30.

You can figure the cost of each for x=30

1.5(30) = 45

1.1(30)+11.5 = 33+11.5 = 44.5 . . . . . lowest cost

1.25(30)+8 = 37.5+8 = 45.5

Speedy Taxis is the cheapest for a trip of 30 miles.

6 0

4 years ago

Tacy deposited $6900 into a savings account that earns 3.1% simple interest

Kisachek [45]

Answer:

$8825.10

Step-by-step explanation:

Interest total= P x r x t

P = principal amount, or $6900

r= interest rate, or 0.031

t = time, or 9 years

6900 x 0.031 x 9 = 1925.10

You now add this to the original amount

6900 + 1925.10 = $8825.10

6 0

4 years ago

I need help I don’t get it

Ahat [919]

Answer:

5 and 9

Step-by-step explanation:

3 multiplied by whatever z is plus 2 has to be 10 or more

6 0

3 years ago

Read 2 more answers

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

6 0

3 years ago

What is the solution to this equation? ​

What is the solution to this equation?