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3 years ago

How many solutions does the equation have?

Mathematics

2 answers:

fgiga [73]3 years ago

7 0

Answer:

A. 0

Step-by-step explanation:

We have been given two equations and we are asked to find the number of solutions our equations have.

$4(2n-4)+3=8n-19$

Let us simplify left side of our equation by distributing 4.

$8n-16+3=8n-19$

$8n-13=8n-19$

We can see that both of our equations have same slope that is 8, but their y-intercepts are different. This means that our equations are parallel and they will never intersect, therefore, our equations will have no solution and option A is the correct choice.

SSSSS [86.1K]3 years ago

6 0

Let's solve and find out.

4(2n - 4) + 3 = 8n - 19
8n - 16 + 3 = 8n - 19
8n - 13 = 8n - 19
-13 = -19

-13 and -19 are not equal, so the equation has no solutions.

Answer:
A. 0

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Natalija [7]

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krek1111 [17]

x = number of 1-cent stamps

y = number of 8-cent stamps

z = number of 12-cent stamps

We have 31 stamps all together, so x+y+z = 31.

"I have 4 more 1-cent stamps than 8-cent stamps" means we have the equation x = y+8. Whatever y is, add 8 to it to get x. Solve for y to get y = x-8.

You also have "twice as many one cent stamps as 12 cent stamps", so x = 2z. Solving for z gets you z = 0.5x

-------------

x+y+z = 31

x+x-8+z = 31 ... y replaced with x-8

x+x-8+0.5x = 31 ... plug in z = 0.5x

2.5x-8 = 31

2.5x = 31+8

2.5x = 39

x = 39/2.5

x = 15.6

Your teacher made a typo somewhere because we should get a positive whole number result for x (since x is a count of how many 1-cent stamps we have).

8 0

3 years ago

A car rents for $250 per week plus $0.50 per mile.find the rental cost for a two-week trip of 500 miles for a group of three peo

Artyom0805 [142]

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Find the sum of a finite geometric sequence from n=1 to n=8, using the expression -2(3)^n-1

MissTica

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}
\end{cases}
\\\\\\
\sum\limits_{n=1}^{8}~-2(3)^{n-1}~~
\begin{cases}
n=8\\
a_1=-2\\
r=3
\end{cases}\implies S_8=-2\left( \cfrac{1-3^8}{1-3} \right)
\\\\\\
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Answer:

Step-by-step explanation

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