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3 years ago

Solve by substitution. X = y + 3 3x + y = 25

1 answer:

3 0

1.distribute
3y+9+y=25
2. combine like terms
4y+9=25
3. subtract 9 from both sides
4y=16
4. divide
Solution: y=4
and for X it would be
X= (4) +3
X=7
so the Final solution is (7,4)

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Isaac shoveled 9 of his neighbors' driveways after a snow storm. He charged $25 for a short driveway and $45 for a long driveway

e-lub [12.9K]

Answer:

y($) = 405- 20x

Step-by-step explanation:

Number of short driveways = x

Number of long driveways = 9-x

Fee per long driveway = $45

Fee per short driveway= $25

Total amount = y

y ($) = 25x + 45 (9-x)

= 25x + 405 -45x

= 405- 20x

5 0

3 years ago

The greatest common factor of 18x -30+12y

12345 [234]

Answer: The greatest common factor is 6.

Step-by-step explanation:

18 goes into 6, 3 times

30 goes into 6, 5 times and q2 goes into 6, 2 times

Divide all three numbers by 6.

The new equation will look like: 3x-5+2y

3 0

3 years ago

Let n = the number. A number exceeds 45 inequality

blondinia [14]

This translates to "a number" is greater than 45. All you have to do now is translate these words into an algebraic statement. Basically, you replace "a number" with the variable which it is defined for, and you use the "greater than" symbol to show that the variable is greater than the value of 45.

3 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$