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uysha [10]
2 years ago
10

Question 5 of 6Which exponential expression is equivalent to the one below?(22• (-7))40A. 40 • (22• (-7))O B. (22) • (-7)40C. (2

2)40 + (-7) 40D. (22)40 . (-7)40+SUBMIT

Mathematics
1 answer:
zmey [24]2 years ago
6 0

Okay, here we have this:

Considering the provided expression, we are going to analize which exponential expression is equivalent, so we obtain the following:

As the property of the exponent of a multiplication says that it is equal to the product of each number raised to that power. We have this:

\begin{gathered} \mleft(22\cdot\mleft(-7\mright)\mright)^{40} \\ =(22)^{40}\cdot(-7)^{40} \end{gathered}

Finally we obtain that the correct answer is the option D.

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A. X= 76, y = 58, z = 104
Alinara [238K]

Answer:

I think A. I'm sorry if I'm wrong

5 0
3 years ago
f r(x) = 3x – 1 and s(x) = 2x + 1, which expression is equivalent to (StartFraction r Over s EndFraction) (6)?
Leya [2.2K]

The equivalent expression to \mathbf{(\dfrac{r}{s})(6)= \dfrac{17}{13}}

<h3>What are the equivalent expressions?</h3>

Equivalent expressions are similar expressions that may have different forms but when a value is replaced with the variables in the same form provides the same answer.

From the information given, If:

  • r(x) = 3x - 1
  • s(x) = 2x + 1

Then the fractional form can be an expression as:

\mathbf{(\dfrac{r}{s})(x)= \dfrac{3x-1}{2x+1}}

where;

  • x = 6

\mathbf{(\dfrac{r}{s})(6)= \dfrac{3(6)-1}{2(6)+1}}

\mathbf{(\dfrac{r}{s})(6)= \dfrac{18-1}{12+1}}

\mathbf{(\dfrac{r}{s})(6)= \dfrac{17}{13}}

Learn more about equivalent expressions here:

brainly.com/question/24734894

#SPJ1

8 0
2 years ago
Which expression is equivalent to 2^5?
N76 [4]

Answer: 2 x 2 x 2 x 2 x 2

Step-by-step explanation:

This is an exponent so you have the base number which is two and the 5 is how much times to multiply the base number.

Hope that helps!

3 0
3 years ago
Four circles of unit radius are drawn with centers $(1,0)$, $(-1,0)$, $(0,1)$, and $(0,-1)$. a circle with radius 2 is drawn wit
Reika [66]
Check the picture 1.

The regions contained in an odd number of these 5 circles are

the pink colored region, contained only in the circle with radius 2, 

and the yellow regions, each contained in the intersection of 2 small circles and the third large circle.

So what we need to determine is the "area pink + area yellow".


Consider the second picture.

Let A2 be the area of the right isosceles triangle, and A1 be half of the yellow region.

the region A2 + A1 is called a sector, and it is 1/4 of the area of a unit circle.

A1=A_{sector}-A2= \frac{1}{4} \pi r^2- \frac{1}{2}.1.1= \frac{1}{4} \pi 1^2- \frac{1}{2}=  \frac{1}{4}\pi- \frac{1}{2}


Thus, the overall yellow region is 8A1=8(\frac{1}{4}\pi- \frac{1}{2})=2\pi-4


The purple area is:
 
Area of large circle - 4*(Area of 1 small circle) + yellow region.

(because subtracting all 4 small circles means that each of the 4 separate yellow regions have been subtracted twice)

A_{purple}=\pi.2^2-4(\pi.1^2)+2\pi-4=4\pi-4\pi+2\pi-4=2\pi-4


Finally, the total area is : 2\pi-4+2\pi-4=4\pi-8 (units squared) 



Answer:   6\pi-8    square units

6 0
4 years ago
The foci and the vertices of the hyperbola are labeled.<br> Which equation represents the hyperbola?
coldgirl [10]

Answer:

Correct option: first one -> (x+2)^2/64 - (y+4)^2/36 = 1

Step-by-step explanation:

The equation of the horizontal major axis hyperbola is:

(x-h)^2/a^2 - (y-k)^2/b^2 = 1

The center is located at (h,k), the vertices are (h+a,k) and (h-a,k) and the focuses are (h+c, k) and (h-c, k)

In this case, the vertices are (-10,-4) and (6,-4), so we have k = -4.

To find h and a, we have:

h+a = 6

h-a = -10

summing both equations, we have:

2h = -4

h = -2

and then for 'a':

-2+a = 6

a = 8

The focus it two units away from the vertix, so c = a + 2, then c = 10

To find b, we can use the relation c^2 = a^2 +b^2:

10^2 = 8^2 + b^2

b^2 = 100 - 64 = 36

b = 6

So the equation of the hyperbola is:

(x+2)^2/64 - (y+4)^2/36 = 1

Correct option: first one

8 0
3 years ago
Read 2 more answers
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