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3 years ago

A. Line a B. Line b C. Line c D. Line d

Mathematics

2 answers:

Cerrena [4.2K]3 years ago

7 0

C. Line c

because it splits the triangle in half.

kotykmax [81]3 years ago

4 0

A line of symmetry is where you can put a line through an object, fold it on that line and it would be the same on both sides. With this triangle it would need to be divided vertically in half therefore it is line c

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I need help with these problems can someone help me!

Tanya [424]

Answer:

1. 14 - 4= 10 2. 8- (-2) = 6 3. 4 - 16 = -12 4. -6 - 3 = -9 5. 13 + (-8) = 5 6. 8 - (-6) = 14

Step-by-step explanation:

5 0

4 years ago

123

ser-zykov [4K]

First, I'm going to separate factor the expression inside of the square root.

sqrt[ (2/18) * (x^5) ]

sqrt[ (1/9) * (x^5) ]

We can take the square root of 1/9 easily, because 1 and 9 are both perfect squares. The square root of 1/9 is 1/3.

Looking at the x^5, we can separate it into x^2 * x^2 * x^1. The square root of x^2 is x. So, we now also have an x^2 on the outside because there are two x^2s in our expanded form.

ANSWER: (x^2 * sqrt(x)) / 3

(Option 1)

Hope this helps!

4 0

3 years ago

A, B and C are events such that P(A) = 1/3, P(B) = ¼, and P(C) = 1/5. Find P(AUBUC) under each of the following assumptions: a)

marta [7]

(a) If $A,B,C$ are mutually exclusive, then

$P(A\cap B)=P(A\cap C)=P(B\cap C)=P(A\cap B\cap C)=0$

so we have

$P(A\cup B\cup C)=P(A)+P(B)+P(C)=\dfrac{47}{60}$

(b) If $A,B,C$ are mutually independent, then

$P(A\cap B)=P(A)P(B),$

$P(A\cap C)=P(A)P(C),$

$P(B\cap C)=P(B)P(C),$

$P(A\cap B\cap C)=P(A)P(B)P(C)$

so that

$P(A\cup B\cup C)=P(A)+P(B)+P(C)-(P(A\cap B)+P(A\cap B)+P(B\cap C))+P(A\cap B\cap C)$

$P(A\cup B\cup C)=\dfrac{47}{60}-\left(\dfrac1{12}+\dfrac1{15}+\dfrac1{20}\right)+\dfrac1{60}$

$P(A\cup B\cup C)=\dfrac35$

5 0

4 years ago

The number 356 is between which pair of numbers? a. 340 and 350 b. 350 and 360 c. 360 and 370 d. 370 and 380

Leno4ka [110]

B. 350, 351, 352, 353, 354, 355, ‘356’, 357, 358, 259, 360

7 0

3 years ago

If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt

MatroZZZ [7]

Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

$\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t$
$\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$

with $0\le t\le1$ . Then we find the differential:

$\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$
$\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$

with $0\le t\le1$ .

Here, the line integral is

$\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle$
$=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$
$=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt$
$=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt$
$=x_1y_2-x_2y_1$

as required.

4 0

4 years ago