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3 years ago

the length of a rectangle is 5 inches more than the width the area is 33 square inches find the length and width

Mathematics

1 answer:

TEA [102]3 years ago

5 0

Try to draw a rectangle and follow those instructions and do u know length times width

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Max is organizing a trip to the airport for a

LuckyWell [14K]

Answer:

$738

Step-by-step explanation:

Uber X costs $40

Max orders 9

$40.00·9=$360.00

Uber XL costs $63

Max orders 6

$63.00·6=$378

$360.00+$378.00=$738

Hope this is helpful :-)

3 0

3 years ago

Find the distance between the two points rounding to the nearest tenth (if necessary).

Evgesh-ka [11]

Answer:

approximately 8.5

Step-by-step explanation:

So you can use the distance formula: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\$ which is derived from the Pythagorean theorem because normally x2-x1 would kind of give the distance length between x2 and x1 except sometimes it would be negative which wouldn't make much sense right? but it doesn't matter because it's being squared so the value ultimately becomes positive, and the same thing goes for the y-value. the distance between x2 - x1 really represents the base and y2-y1 represents the height and then adding them together gives the hypotenuse squared which is why you take the square root over the entire thing to find the distance, since the hypotenuse is the shortest distance between two points

Now all you have to do is plug the values in to get

$\sqrt{(-7 - (-4))^2 + (-8 - 0)^2}\\\sqrt{(-3)^2 + (-8)^2}\\\sqrt{9 + 64}\\\sqrt{73}\\distance \approx 8.5$

5 0

2 years ago

6n + n^7 is divisible by 7 and prove it in mathematical induction

kompoz [17]

Answer:

Apply induction on $n$ (for integers $n \ge 1$ ) after showing that $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7$ for $j \in \lbrace 1,\, \dots,\, 6 \rbrace$ .

Step-by-step explanation:

Lemma: $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7$ for $j \in \lbrace 1,\, 2,\, \dots,\, 6\rbrace$ .

Proof: assume that for some $j \in \lbrace 1,\, 2,\, \dots,\, 6\rbrace$ , $\genfrac{(}{)}{0}{}{7}{j}$ is not divisible by $7$ .

The combination $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is known to be an integer. Rewrite the factorial $7!$ to obtain:

$\displaystyle \begin{pmatrix}7 \\ j\end{pmatrix} = \frac{7!}{j! \, (7 - j)!} = \frac{7 \times 6!}{j!\, (7 - j)!}$ .

Note that $7$ (a prime number) is in the numerator of this expression for $\genfrac{(}{)}{0}{}{7}{j}\!$ . Since all terms in this fraction are integers, the only way for $\genfrac{(}{)}{0}{}{7}{j}$ to be non-divisible by $7\!$ is for the denominator $j! \, (7 - j)!$ of this expression to be an integer multiple of $7\!\!$ .

However, since $1 \le j \le 6$ , the prime number $\!7$ would not a factor of $j!$ . Similarly, since $1 \le 7 - j \le 6$ , the prime number $7\!$ would not be a factor of $(7 - j)!$ , either. Thus, $j! \, (7 - j)!$ would not be an integer multiple of the prime number $7$ . Contradiction.

Proof of the original statement:

Base case: $n = 1$ . Indeed $6 \times 1 + 1^{7} = 7$ is divisible by $7$ .

Induction step: assume that for some integer $n \ge 1$ , $(6\, n + n^{7})$ is divisible by $7$ . Need to show that $(6\, (n + 1) + (n + 1)^{7})$ is also divisible by $7\!$ .

Fact (derived from the binomial theorem $(\ast)$ ):

$\begin{aligned} & (n + 1)^{7} \\ &= \sum\limits_{j = 0}^{7} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right] && (\ast)\\ &= \genfrac{(}{)}{0}{}{7}{0} \, n^{0} + \genfrac{(}{)}{0}{}{7}{7} \, n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right] \\ &= 1 + n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\end{aligned}$ .

Rewrite $(6\, (n + 1) + (n + 1)^{7})$ using this fact:

$\begin{aligned} & 6\, (n + 1) + (n + 1)^{7} \\ =\; & 6\, (n + 1) + \left(1 + n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right) \\ =\; & 6\, n + n^{7} + 7 + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right) \end{aligned}$ .

For this particular $n$ , $(6\, n + n^{7})$ is divisible by $7$ by the induction hypothesis.

$\sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]$ is also divisible by $7$ since $n$ is an integer and (by lemma) each of the coefficients $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7\!$ .

Therefore, $6\, (n + 1) + (n + 1)^{7}$ , which is equal to $6\, n + n^{7} + 7 + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right)$ , is divisible by $7$ .

In other words, for any integer $n \ge 1$ , if $(6\, n + n^{7})$ is divisible by $7$ , then $6\, (n + 1) + (n + 1)^{7}$ would also be divisible by $7\!$ .

Therefore, $(6\, n + n^{7})$ is divisible by $7$ for all integers $n \ge 1$ .

6 0

2 years ago

HELP ANSWER /////////////////////////////////////////////////////////////////

stepan [7]

Answer:

Step-by-step explanation:

$4 \times 6 \times 4 = 96cubic \: units$

8 0

3 years ago

Yuri measures the height of water in an aquarium in one metric unit. She then moves the decimal point 3 units to the left. Which

irakobra [83]

Answer: A) Kiloliters to liters

If the tank was say 1000 kiloliters, then that is equivalent to 1 liter according to the chart. We divide by 1000 to go from kiloliters to liters, which is the same as moving the decimal point 3 spots to the left.

7 0

3 years ago

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