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2 years ago

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The lenngths of the sides of the triangle are in the extended ratio 8:9:10. The primeter of the triangle is 108 cm.What are the

lengths of the sides?
The lengths of the sides of the triangle in cm are____?
Please help I really need this!

1 answer:

hjlf2 years ago

4 0

Answer:

Side known in ratio as 8: 32

Side known in ratio as 9: 36

Side known in ration as 10: 40

Step-by-step explanation:

We can add all the numbers in the ration to form the sum of all the sides, or also equal to perimeter. 8+9+10 equals 27.

So we can write 27x=108

x=4. Multiply 4 to 8, then 9, then 10. You get 32, 36, 40

Hope this helped, Have a Great Day!!

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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Evaluate the expression below using the properties of operations. Show your steps.−36 ÷ 14 ⋅ (−18) ⋅ (−3) ÷ 6

s2008m [1.1K]

Answer:

(-162)/7 or -23 1/7 as mixed fraction

Step-by-step explanation:

Simplify the following:

(-36)/14 (-18) (-3)/6

Hint: | Express (-36)/14 (-18) (-3)/6 as a single fraction.

(-36)/14 (-18) (-3)/6 = (-36 (-18) (-3))/(14×6):

(-36 (-18) (-3))/(14×6)

Hint: | In (-36 (-18) (-3))/(14×6), divide -18 in the numerator by 6 in the denominator.

(-18)/6 = (6 (-3))/6 = -3:

(-36-3 (-3))/14

Hint: | In (-36 (-3) (-3))/14, the numbers -36 in the numerator and 14 in the denominator have gcd greater than one.

The gcd of -36 and 14 is 2, so (-36 (-3) (-3))/14 = ((2 (-18)) (-3) (-3))/(2×7) = 2/2×(-18 (-3) (-3))/7 = (-18 (-3) (-3))/7:

(-18 (-3) (-3))/7

Hint: | Multiply -18 and -3 together.

-18 (-3) = 54:

(54 (-3))/7

Hint: | Multiply 54 and -3 together.

54 (-3) = -162:

Answer: (-162)/7

3 0

3 years ago

A bus travels on an east-west highway connecting two cities A and B that are 100 miles apart. There are 2 services stations alon

melamori03 [73]

Answer:

51/4

Step-by-step explanation:

To begin with you have to understand what is the distribution of the random variable. If X represents the point where the bus breaks down. That is correct.

X~ Uniform(0,100)

Then the probability mass function is given as follows.

$f(x) = P(X=x) = 1/100 \,\,\,\, \text{if} \,\,\,\, 0 \leq x \leq 100\\f(x) = P(X=x) = 0 \,\,\,\, \text{otherwise}$

Now, imagine that the D represents the distance from the break down point to the nearest station. Think about this, the first service station is 20 meters away from city A, and the second station is located 70 meters away from city A then the mid point between 20 and 70 is (70+20)/2 = 45 then we can represent D as follows

$D(x) =\left\{ \begin{array}{ll} x & \mbox{if } 0\leq x \leq 20 \\ x-20 & \mbox{if } 20\leq x < 45\\ 70-x & \mbox{if } 45 \leq x \leq 70\\ x-70 & \mbox{if } 70 \leq x \leq 100\\ \end{array}\right.$

Now, as we said before X represents the random variable where the bus breaks down, then we form a new random variable $Y = D(X)$ , $Y$ is a random variable as well, remember that there is a theorem that says that

$E[Y] = E[D(X)] = \int\limits_{-\infty}^{\infty} D(x) f(x) \,\, dx$

Where $f(x)$ is the probability mass function of X. Using the information of our problem

$E[Y] = \int\limits_{-\infty}^{\infty} D(x)f(x) dx \\= \frac{1}{100} \bigg[ \int\limits_{0}^{20} x dx +\int\limits_{20}^{45} (x-20) dx +\int\limits_{45}^{70} (70-x) dx +\int\limits_{70}^{100} (x-70) dx \bigg]\\= \frac{51}{4} = 12.75$

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3 years ago

Question Navigator

MArishka [77]

C and B are correct,

8 0

2 years ago

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garik1379 [7]

Answer:

NOPE

Step-by-step explanation:

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2 years ago