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mina [271]
3 years ago
7

Determine whether each of the following triangles is a right-angled triangle

Mathematics
1 answer:
Fantom [35]3 years ago
4 0

Answer:

Yes

Step-by-step explanation:

To determine if this is a right triangle, use the pythagorean theorem and see if a^2 + b^2 is equal to c^2.

a^2 + b^2 = c^2

12^2 + 16^2 = 20^2

Evaluate the exponents.

144 + 256 = 400

Add them together.

400 = 400

Since a^2 and b^2 add up to equal c^2, this is a right triangle.

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The polynomial 16x +2 is a ? <br> Adding the term ? Would make it a trinomial
trapecia [35]

What are the instructions? If it just says to classify it, there is no degree, so it would be a binomial.

Here is a chart;

1 term = monomial. Example; 6x

2 terms = binomial. Example; 6x + 5

3 terms = Trinomial. Example; 6x + 5 + 2y

4 terms = polynomial of 4 terms. Example; 6x + 5 + 2y + 2x^3

7 0
3 years ago
Read 2 more answers
Please assist me with these problems using the 45-45-90 relationship triangles​
TiliK225 [7]

Answer:

see below

Step-by-step explanation:

side opposite 90 angle = x \sqrt{2}

sides opposite 45 angle = x

the multiplication and division by  \sqrt{2} should be self explanatory

1. x = 5, y = 5

2. v = 10, u = 10

3. x = \sqrt{10}, y = \sqrt{5}

4. m = 5, n = \frac{5\sqrt{2} }{2}

5. a = \frac{3\sqrt{2} }{2}, b =  \frac{3\sqrt{2} }{2}

6. a = \frac{3\sqrt{2} }{2}, b =  \frac{3\sqrt{2} }{2}

5 0
3 years ago
How do you graph <br> y=3x and y=-x-8
laila [671]

Answer:

The points for the given two linear equation as

x_1 , y_1 = - 2 , - 6

x_2 , y_2 = - 2 , 6

The graph so plotted as shown

Step-by-step explanation:

Given as :

The two linear equation are

y = 3 x                 ........A and

y = - x - 8             .........B

Solving equation A and B

Now, Put The value of y from eq A into eq B

So, 3 x = - x - 8

Or, 3 x + x = - 8

Or, 4 x = - 8

∴ x = \dfrac{-8}{4}

I.e x = - 2

Now , Put the value of x into eq A

∵ y = 3 x

∴ y = 3 × (-2)

I.e y = - 6

Again, Put the value of x into eq B

∵ y = - x - 8

∴ y = - 2 - (-8)

I.e y = 6

So, for x = - 2 , y = - 6

And for x = - 2 , y = 6

Hence , The points for the given two linear equation as

x_1 , y_1 = - 2 , - 6

x_2 , y_2 = - 2 , 6

The graph so plotted as shown . Answer

8 0
3 years ago
The minimum temperature for a particular day was 4 Fahrenheit below 0 the maximum was 16 Fahrenheit higher what was the maximum
Talja [164]
16 degrees would be the maximum because it was the hottest temperature recorded.
5 0
3 years ago
A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat
zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

Z = \frac{X - \mu}{s}

X = 205

Z = \frac{X - \mu}{s}

Z = \frac{205 - 200}{5}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 195

Z = \frac{X - \mu}{s}

Z = \frac{195 - 200}{5}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

Z = \frac{X - \mu}{s}

Z = \frac{210 - 200}{5}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 195

Z = \frac{X - \mu}{s}

Z = \frac{190 - 200}{5}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

7 0
3 years ago
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