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3 years ago

The hypotenuse of a right triangle is 13 cm. one of the legs is 7 cm longer than the other leg. find the area of the triangle.

1 answer:

8 0

Let one side be=x,the longer side =x+7use pythagoras theoremH^2=b^2+h^2169=x^2+(x+7)^2169=2x^2+14x+490=2x^2+14x-1200=(2x-10)(x+12) x=5 or -12,since it is length x=5 the longer side=5+7=12confirmation169=144+25

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Answer:

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Step-by-step explanation:

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3 years ago

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Please help 14 points ASAP

grigory [225]

The desired measures for the data-set is given by:

Minimum: 48

Lower quartile: 54.

Median: 63.5.

Upper quartile: 74.

Maximum: 80.

IQR: 20

<h3>How to find the five number summary and interquartile range of the data-set?</h3>

The five number summary is composed by the measures explained below, except the IQR.

The minimum value is the smallest value from the data-set, as the maximum value is the greatest value of the data-set.

The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.

The first quartile is the median of the first half of the data-set.

The third quartile is the median of the second half of the data-set.

The interquartile range is the difference of the third quartile and the first quartile.

In this problem, we have that:

The minimum value is the smallest value, of 48.

The maximum value is the smallest value, of 80.

The data-set has even cardinality, hence the median is the mean of the middle elements, which are 63 and 64, hence the median is of 63.5.

The first quartile is the median of the five elements of the first half, hence it is of 54.

The third quartile is the median of the five elements of the second half, hence it is of 74.

The IQR is the difference between the quartiles, hence 74 - 54 = 20.

More can be learned about five number summaries at brainly.com/question/17110151

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2 years ago

Please help me with this stuff, thanks.

zysi [14]

This is really ominous

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3 years ago

Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x

soldier1979 [14.2K]

Answer:

(a) The function is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is $x = -\frac{1}{2}$

(d) The function is concave upward on $\left(- \frac{1}{2}, \infty\right)$ and concave downward on $\left(-\infty, - \frac{1}{2}\right)$

Step-by-step explanation:

(a) To find the intervals where $f(x) = 2x^3 + 3x^2 -180x$ is increasing or decreasing you must:

1. Differentiate the function

$\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180$

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

$f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6$

These points divide the number line into three intervals:

$(-\infty,-6)$ , $(-6,5)$ , and $(5, \infty)$

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

$\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right$

Therefore f(x) is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.

f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

$f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6$

We set f''(x) = 0

$f''(x) =12x+6 =0\\\\x=-\frac{1}{2}$

Analyzing concavity, we get

$\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right$

The function is concave upward on $(-1/2,\infty)$ because the f''(x) > 0 and concave downward on $(-\infty,-1/2)$ because the f''(x) < 0.

f(x) is concave down before $x = -\frac{1}{2}$ , concave up after it. So f(x) has an inflection point at $x = -\frac{1}{2}$ .

7 0

3 years ago