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3 years ago

Which set of sides will make a triangle?

Mathematics

1 answer:

Klio2033 [76]3 years ago

6 0

Answer:

10cm 5cm and 9cm

Step-by-step explanation:

just try this

take any two measurements and add them together if they are larger than the 3rd measurement it will work and all you have to do is do that to each measurement

10+5 is greater than 9

9+5 is greater than 10

9+10 is greater than 5

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The value of N is both 5 times as much as the value of M and 36 more than the value of M. What are the values of N and M? Explai

vlabodo [156]

180 just multiply 36*5= 180

Thank you, i would be happy to have more questions

7 0

3 years ago

How do I solve this, and what is the answer? please and thank you.

vovangra [49]

Answer:

A. (x +8) + (-4x+31)/(x^2+2x+1)

Step-by-step explanation:

When you perform long division of polynomials, the first quotient term is the ratio of the highest-degree terms in the numerator and denominator: x^3/x^2 = x.

This fact eliminates all but choices A and C.

The denominator of the remainder term is the denominator of the original expression, so will be x^2 +2x +1, as shown in choice A.

So, simply based on a couple of facts about long division (that you learned in the early elementary grades), you can make the correct choice of answer without actually working the problem in detail.

_____

This is a polynomial long division problem. It is worked in virtually the same way that numerical long division problems are worked: first you find a quotient term, then you multiply that by the divisor and subtract the result from the dividend. The difference is the new dividend. These steps are identical to numerical long division.

For polynomial long division, instead of lining up the digits with the same place value, you line up the terms with the same degree of the variable.

As mentioned above, the quotient term is computed only from the highest-degree terms of dividend and divisor, so that part is actually simpler than for numerical long division.

A dividend that is of lower degree than the divisor is considered to be the remainder. As with numerical long division, it can be expressed as a fraction with the divisor as the denominator.

Numerical example: 18/7 = 2 remainder 4 or 2 4/7.

8 0

3 years ago

Luke earned $36,000 during the first year of his job at Putt-Putt Kingdom. After each year he received a 10% raise. Find his tot

Lady_Fox [76]

In order to solve this problem, you need to use a geometric series:
$S_{n} = \frac{ a_{1}(1 - r^{n}) }{1 - r}$
where:
a₁ = first term of the series = 36000
r = common rate = 10% raise, therefore 1.10
n = number of terms = 5

Therefore,
<span> $S_{n} = \frac{ 36000(1 - 1.10^{5}) }{1 - 1.10}$
= 219783.60 $

Luke's total earnings in five years are <span>219783.60 $.</span>

</span>

6 0

3 years ago

To test the effectiveness of a business school preparation course, 8 students took a general business test before and after the

Lisa [10]

Solution :

Group Before After

Mean 693.75 743.75

Sd 155.37 143.92

SEM 54.93 50.88

n 8 8

Null hypothesis : The preparation course not effective.

$H_0: \mu_d = 0$

Alternative hypothesis : The preparation course is effective in improving the exam scores.

$H_a : \mu_d>0$ (after - before)

4 0

3 years ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 6

$\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28$

When r = 7

$\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 8

$\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1$

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0

3 years ago