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2 years ago

-5/6x = 20 2/3(x - 7) = -2 solve them and show your work! ASAP! I need this done fast!!

Mathematics

1 answer:

LenKa [72]2 years ago

3 0

Answer:

x=-24 & x=4.

Step-by-step explanation:

-5/6x=20

x=20/(-5/6)

x=(20/1)(-6/5)=-120/5=-24

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

2/3(x-7)=-2

2/3x-14/3=-2

2/3x=-2+14/3

2/3x=-6/3+14/3

2/3x=8/3

x=(8/3)/(2/3)

x=(8/3)(3/2)=8/2=4

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in how many ways can a student arrange 6 textbooks on a locker shelf that can hold 4 books at a time? 30 120 240 360

denis23 [38]

On the 1st place he can put 6 textbooks, on the 2nd 5 textbooks etc...
6 · 5 · 4 · 3 = 360
Answer: D) He can arrange textbooks in 360 ways.

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3 years ago

Which expression represents the area of the shaded region?

irina1246 [14]

Given:

A circle of radius r inscribed in a square.

To find:

The expression for the area of the shaded region.

Solution:

Area of a circle is:

$A_1=\pi r^2$

Where, r is the radius of the circle.

Area of a square is:

$Area=a^2$

Where, a is the side of the square.

A circle of radius r inscribed in a square. So, diameter of the circle is equal to the side of the square.

$a=2a$

So, the area of the square is:

$A_2=(2r)^2$

$A_2=4r^2$

Now, the area of the shaded region is the difference between the area of the square and the area of the circle.

$A=A_2-A_1$

$A=4r^2-\pi r^2$

$A=r^2(4-\pi )$

Therefore, the correct option is (a).

5 0

3 years ago

What is the most difficult math problem in the world?

solmaris [256]

xn+yn=zn where n>2 and n is an integer

<h2>bolded letters are exponents!!!!</h2>

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3 years ago

Read 2 more answers

I don’t understand what this question is asking for

Elden [556K]

Answer:

(d) m∠AEB = m∠ADB

Step-by-step explanation:

The question is asking you to compare the measures of two inscribed angles. Each of the inscribed angles intercepts the circle at points A and B, which are the endpoints of a diameter.

<h3>applicable relations</h3>

Several relations are involved here.

The measures of the arcs of a circle total 360°
A diameter cuts a circle into two congruent semicircles
The measure of an inscribed angle is half the measure of the arc it intercepts

<h3>application</h3>

In the attached diagram, we have shown inscribed angle ADB in blue. The semicircular arc it intercepts is also shown in blue. A semicircle is half a circle, so its arc measure is half of 360°. Arc AEB is 180°. That means inscribed angle ADB measures half of 180°, or 90°. (It is shown as a right angle on the diagram.)

If Brenda draws angle AEB, it would look like the angle shown in red on the diagram. It intercepts semicircular arc ADB, which has a measure of 180°. So, angle AEB will be half that, or 180°/2 = 90°.

The question is asking you to recognize that ∠ADB = 90° and ∠AEB = 90° have the same measure.

m∠AEB = m∠ADB

_____

<em>Additional comment</em>

Every angle inscribed in a semicircle is a right angle. The center of the semicircle is the midpoint of the hypotenuse of the right triangle. This fact turns out to be useful in many ways.

7 0

2 years ago

Evaluate the indefinite integral as a power series. t 1 − t7 dt

sammy [17]

Given

$\int\limits { \frac{t}{1-t^7} } \, dt$

Recall that the general geometric series representation is given by

$\Sigma_0^\infty x^k= \frac{1}{1-x}$

Let $x=t^7$

Thus,

$\frac{1}{1-t^7}=\Sigma_0^\infty (t^7)^k=\Sigma_0^\infty t^{7k}$

Multiply both sides by t, to get

$\frac{1}{1-t^7}=\Sigma_0^\infty t^{7k+1}$

Integrate both sides to get

$\int { \frac{t}{1-t^7} } \, dt=\Sigma_0^\infty \frac{t^{7k+2}}{7k+2}$

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2 years ago