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ivanzaharov [21]
3 years ago
9

Please help with the question on the picture

Mathematics
1 answer:
Advocard [28]3 years ago
4 0

Answer:

c) = (-180)

d) = (-81)

this is the answer that is to be written in the boxes

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(D) \frac{\sqrt{x}}{x}

Step-by-step explanation:

\frac{3x^2}{\sqrt{9x^5}}=\frac{3x^2}{3\sqrt{x^5}}=\frac{\sqrt{x^4}}{\sqrt{x^5}}=\sqrt{\frac{x^4}{x^5}}=\\=\sqrt{\frac{1}{x}}=\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}

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1 x 10^12 = 9 x 10^8 (8/d^2) solve for d
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4 years ago
the length of a rectangle exceeds its width by 6 inches and the area is 40 square inches. What are the length and width of the r
Lorico [155]

Classic Algebra and its unnecessarily complicated sentence structure. As you may have probably known, Algebra has its own "vocabulary set".


"the length of a rectangle exceeds its width by 6 inches" -> length is 6 in. longer than width -> l= w + 6


Since we're solving for the length and width, let's give them each variables.

length = l = w+6

width = w


The next bit of information is "the area is 40 square inches"

Applying the formula for the area of a rectangle we can set up:

l x w = 40

replace "l", or length, with it's alternate value.

(w+6) x w = 40

distribute

w^{2} + 6w = 40

subtract 40 from both sides

w^{2} + 6w - 40 = 0

factor

(w - 4)(w + 10) = 0

solve for w

w= 4, or -10


So great, we have 2 values; which one do we choose? Since this problem is referring to lengths and inches, we will have to choose the positive value. There is not such thing as a negative distance in the real world.


We now have half of the problem solved: width. Now we just need to find the length which we can do but substituting it back into the original alternate value of l.


l = w + 6

w=4

l = 4 + 6 = 10


The length is 10 in. and the width is 4 in. Hope this helps!

8 0
4 years ago
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