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Margarita [4]
3 years ago
14

Right triangle XYZ has legs of length XY = 12 and YZ = 6. Point D is chosen at random within the triangle XYZ. What is the proba

bility that the area of triangle XYD is at most 12?
Mathematics
2 answers:
Llana [10]3 years ago
8 0

The "successful" region has half the size of the whole triangle, so by similar triangles, the probability that the area of triangle XYD is at most 12 is 1/4.

Brilliant_brown [7]3 years ago
7 0

Answer:

5/9

Step-by-step explanation:

If you draw the triangle out on a grid papar and draw a line across the 2 marker point from XY up of the YZ side, you will see that the line from that marker point across to the hypotenuse is actually 8. Then you find the area of the little triangle with legs 8 and 4 (since 6-2 is 4), and the area is 16. After that, find the total area of the big triangle, which is 12x6/2=36. Since we want to know the area of the quadrilateral at the bottom of the big triangle seperated by the 2 marker point, we subtract 16 from 36, which equals to 20. 20/36=5/9. And that is my final answer. I tested it and it is correct. I hope this helps.

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Use the quadratic model y = −4x2 − 3x + 4 to predict y if x equals 5.
REY [17]

Answer:

-111

Step-by-step explanation:

substitute 5 in place of each x

    -4(5)^2 - 3(5) + 4

then evaluate or type it into your calculator

    -100 - 15 + 4

7 0
3 years ago
Read 2 more answers
Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va
Burka [1]

Answer:

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

The polynomial is an approximation with an error less than or equals to <em>0.002652</em> for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)

with

\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}

for some c in the interval (-x, x)

In the particular case f

<em>f(x)=cos(x) </em>

<em> </em>

we have

\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)

therefore

\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0

and the polynomial approximation of T5(x) of cos(x) would be

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}

for some c in (-x,x). So

\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}

and we must find the values of x for which

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652

Working this inequality out, we find

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0
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timurjin [86]
Are there any answer choices maybe??
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Greeley [361]

Answer:

19 4/5

Step-by-step explanation:

4 0
3 years ago
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