Which of the follow expressions are equivalent to (3^4)^-2 • (3^5)^3

Mathematics

2 answers:

Harrizon [31]3 years ago

4 0

Answer:

3^7

Step-by-step explanation:

(3^4)^-2 • (3^5)^3 = 2187

3^7 = 2187

Fiesta28 [93]3 years ago

3 0

Answer:

3^7

Step-by-step explanation:

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MARK BRAINIEST The base of a triangle is 7 3/4 inches, and the height is 4 2/3 inches. What is the area of the triangle? A) 14 4

Keith_Richards [23]

Short Answer: B
Remark
The area of any triangle is
A = 1/2 b * h
b = 7 3/4
h = 4 2/3

Step one
Convert all fractions to improper fractions.

7 3/4 = (28 + 3) / 4 = 31/ 4

4 2/3 = (12 + 2) / 3 = 14/3 Now multiply

Step Two
Multiply
Area = 1/2 * 31/4 * 14/3 Multiply the numerators together and then the denominators.
Area = (1 * 31 * 14) / (2 * 4 * 3) = 434 /24

Step 3
Convert this to a whole number + a fraction

Area = 18 + 0.08333333 You already have your answer. It is B because only B gives an answer of 18. But you can get the rest of

Let x = 0.083333333
100 x = 8.3333333333
<u>x = 0.0833333333 </u> Subtract
99 x = 8.25 Divide by 3 on both sides.
33 x = 2.75 Divide by 11
3x = 0.25
x = 0.25/3 Multiply top and bottom by 4
x = 1 / 12

The answer is 18 1/12.

3 0

3 years ago

If you answer this you cool but still PLEASEE HELPP

defon

Its D

Because the rate of change is 1.5 and 1.5x10 is 15

5 0

3 years ago

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$