Use partial quotients 6 divided by 258

Which expression is equivalent to 128x^5y^6 \ 2x^7y^5 ? Assume x > 0 and y > 0.

N76 [4]

Answer: Last option.

Step-by-step explanation:

Given the expression:

$\sqrt{\frac{128x^5y^6}{2x^7y^5}$

The Quotient of powers property states that:

$\frac{a^m}{a^n}=a^{(m-n)}$

And the Power of a powet property states that:

$(a^m)^n=a^{mn}$

Then, applying these properties, you get:

$=\sqrt{\frac{(2^3)^26y}{x^2}$

Now you must remember that:

$\sqrt[a]{a^n}=a$

Therefore, simpliying the expression, you get:

$=\frac{2^3\sqrt{y}}{x}=\frac{8\sqrt{y}}{x}$

7 0

3 years ago

Read 2 more answers

Pls help me so I can pass

SashulF [63]

No idea but good luck

8 0

3 years ago

Read 2 more answers

Help please guys thankyouu

ZanzabumX [31]

You have the right idea so far, but you forgot to finish it up.

Point slope form is

y - y1 = m(x - x1)

where m is the slope, and (x1,y1) is the point the line goes through. As you can see, we have infinitely many choices for (x1,y1). If you use the y intercept point (0,5), then

y - y1 = m(x - x1)

y - 5 = -1(x - 0) ..... this is one of infinitely many possible answers

OR

If you use the point (2,3), then we'd say

y - y1 = m(x - x1)

y - 3 = -1(x - 2) ..... which is another possible answer

There are infinitely many other choices, but I'm only picking the points in which your teacher has shown as big dots.

By the way, nice work on getting the correct answer for the slope-intercept form.

8 0

3 years ago

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A point x is 34m due to east of a point Y . The bearing of a flagpole from x and Y are N18°W And N40°E respectively .

Murljashka [212]

Answer:

The distance of flag post from Y is 38.13 m

Step-by-step explanation:

Consider point Y at the intersection of both lines as shown below. Now the point X lies 34 meter from point Y in east direction.

Now flag pole at point X lies at a bearing of N18°W. That is at point X from north, flag post makes an angle of 18° towards west.

Similarly flag pole at point Y lies at a bearing of N40°E. That is at point Y from north, flag post makes an angle of 40° towards east.

Consider ∆ AXY as right angle triangle. Therefore measure of angle FXY is,

$\angle AXY=\angle AXF+\angle FXY$

$\therefore 90 \degree=18\degree+\angle FXY$

$\angle FXY=72\degree$

Consider ∆ BYX as right angle triangle. Therefore measure of angle FYX is,

$\angle BYX=\angle BYF+\angle FYX$

$\therefore 90 \degree=40\degree+\angle FYX$

$\angle FYX=50\degree$

Refer attachment 1.

From diagram consider the triangle FYX. To find the third angle that is ∠YFX can be calculated by using angle sum property of triangle.

∠YFX+∠FYX+∠FXY=180°

∠YFX+50°+72°=180°

∠YFX=58°

Refer attachment 2.

Now the distance FY can be calculated using sin rule as follows,

$\frac{\sin X}{FY}=\frac{\sin F}{YX}=\frac{\sin Y}{FX}$

Substituting the values,

$\frac{\sin 72}{FY}=\frac{\sin 58}{34}=\frac{\sin 50}{FX}$

Simplifying first two terms,

$\frac{\sin 72}{FY}=\frac{\sin 58}{34}$

Cross multiplying,

$34\times\sin 72=FY\times \sin 58$

$\dfrac{34\times\sin 72}{\sin 58}=FY$

$FY=38.13$ m

7 0

3 years ago

HELP PLS!! <br> Adding and subtracting vectors in component form

tino4ka555 [31]

<h3>Answers:</h3>

u+v = <3,12>
w+g = <7,0>
g-z = <2,4>
v-u = <9,4>
y+v = <7,9>
u+v+y = <4,13>

===================================================

Explanation:

Problem 1

If we had the two vectors u = <a,b> and v = <c,d>, then adding them gives us

u+v = <a+c,b+d>

The corresponding coordinates pair up and add together.

In this case we have

u = <-3,4>

v = <6,8>

So,

u+v = <-3+6,4+8>

u+v = <3,12>

---------------------

Problem 2

We follow the same idea as the previous problem.

w = <8,-1>

g = <-1,1>

w+g = <8+(-1),-1+1>

w+g = <7,0>

---------------------

Problem 3

Similar to addition, subtracting vectors has us subtract the corresponding coordinates.

The general template is:

u = <a,b>

v = <c,d>

u-v = <a-c,b-d>

With this in mind, we can say the following:

g = <-1,1>

z = <-3,-3>

g-z = <-1-(-3),1-(-3)>

g-z = <-1+3,1+3>

g-z = <2,4>

---------------------

Problem 4

Follow the same idea as problem 3 above.

v = <6,8>

u = <-3,4>

v-u = <6-(-3),8-4>

v-u = <6+3,8-4>

v-u = <9,4>

---------------------

Problem 5

Refer to problem 1.

y = <1,1>

v = <6,8>

y+v = <1+6,1+8>

y+v = <7,9>

---------------------

Problem 6

u = <-3,4>

h = v+y = y+v = <7,9>

u+v+y = u + h

u+v+y = <-3,4> + <7,9>

u+v+y = <-3+7,4+9>

u+v+y = <4,13>

Notice how I built off the result of problem 5 when I used h = v+y. The vector v+y is the same as y+v because the order of addition doesn't matter. Also, the idea mentioned in problem 1 can be extended for more than two vectors.

3 0

3 years ago