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4 years ago

A triangle has a perimeter of 90 centimeters write and use a linear system

Mathematics

1 answer:

lawyer [7]4 years ago

5 0

We are given sides of the triangle by expression

First side : n

Second side : m = 5n-5.

Third side : l = 5/4 m - n.

Perimeter is 90.

Therefore, first equation could be set :

n +m +l = 90 --------------equation(1)

Second equation would be

m = 5n-5 --------------equation(2)

Third equation would be

l = 5/4 m - n --------------equation(3)

Let us solve the system of equations by substitution.

Substituting m = 5n-5 and l = 5/4 m - n in first equation.

We get

n +5n-5 +5/4 m - n= 90

Now, substituting m = 5n-5 in above equation we get

n +5n-5 +5/4 (5n-5) - n= 90

5n + 25n/4 -5 - 25/4 =90

5n + 6.25n -5-6.25 = 90.

11.25n-11.25 = 90

Adding 11.25 on both sides, we get

11.25n-11.25+11.25 = 90+11.25

11.25n = 101.25.

Dividing both sides by 11.25, we get

n=9.

Plugging n=9 in 2nd equation.

m = 5(9)-5 = 45 -5 = 40.

Plugging n=9 and m=40 in first equation, we get

9 +40 +l = 90

49 +l = 90.

Subtracting 49 from both sides, we get

49-49 +l = 90-49

l = 41.

Therefore, sides are of lengths l = 41, m= 40 and n=9.

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7. Jerry has 23 coins in nickels, n, and dimes, d. He

aliya0001 [1]

Answer:

17 nickles !

Step-by-step explanation:

First, identify the variables:

n = amount of nickels

d = amount of dimes

Next, setup the equations based on what you know. The first equation is:

n + d = 28

For the second equation, we know that a dime is worth 10¢ and a nickel is 5¢, so it should be:

0.05n + 0.10d = 1.95

This a three-step answer:

In one formula (you can use any of them; most people use the simplest one), single out the variable on one side

Apply the first formula into the second formula, and solve it to get the value of one variable

Apply the answer from the second formula into the first formula, and solve it to get the value of the other variable

======

Step One:

n + d = 28

n + d - d = 28 - d

n = 28 - d

Step Two:

0.05n + 0.10d = 1.95

(0.05 * (28 - d)) + 0.10d = 1.95

1.40 - 0.05d + 0.10d = 1.95

1.40 + 0.05d = 1.95

1.40 - 1.40 + 0.05d = 1.95 - 1.40

0.05d = 0.55

d = 11

Step Three:

n = 28 - d

n = 28 - 11

n = 17

======

Your answer should be 17 nickels and 11 dimes.

You can double check by applying the variables into both formulas.

n + d = 28

17 + 11 = 28

28 = 28

0.05n + 0.10d = 1.95

(0.05 * 17) + (0.10 * 11) = 1.95

0.85 + 1.10 = 1.95

1.95 = 1.95

I hope this helped.

7 0

3 years ago

BANK, BA=7cm,BK=5.6cm,and angle KBA=85 degree

ioda

What is the question? What do I do with this information

8 0

3 years ago

93 round to the nearest hundred

klasskru [66]

5 or more add one more
4 or less could care less(aka don't change the number)
93 right and round it to the nearest hundreds
So ig 93 than you look at the tens is it more than 5 or less than 4?
It's more than 5 so you add another number on the hundreds since there is nothing on the hundred place it is 100(Answer)if it was less than 4 it would just be 0. Hope I helped

6 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of