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3 years ago

Please can you help me! z-² as a fraction

1 answer:

5 0

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
z^{-2}\implies \cfrac{1}{z^2}$

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What is the product written in scientific notation? (5.91×10−3)⋅(8.7×1010)

trasher [3.6K]

<span>(5.91×10^−3)⋅(8.7×10^10)
=</span><span>(5.91 × 8.7)⋅(10^-3 × 10^10)
= 51.417 x 10^7
= 5.1417 x 10^8

hope it helps</span>

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4 years ago

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A typical dwarf sperm whale, the planets smallest whale species, weighs about 3*10^2 pounds. A blue whale, the planets largest w

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Answer:

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Step-by-step explanation:

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3 years ago

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ololo11 [35]

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3 years ago

If j=66 , l=64 , k=50 in order from smallest to largest.

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K, I, J. 59, 64, 66.

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4 years ago

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A ladder 16 feet long is leaning against the wall of a tall building. The base of the ladder is moving away from the wall at a r

svet-max [94.6K]

Answer:

a. 0.588

b. 0.0722

c. 4.576 sqft/sec

Step-by-step explanation:

Let b and h denote the base and height as indicated in the diagram. By pythagoras theorem, $h^2 + b^2 = 16^2 = 256 \dotsc\;(1)$ because it is a right angle triangle.

It is given that $\frac{db}{dt} = 1$

Now differentiate (1) with respect to t (time) :

$\displaystyle{2h\frac{dh}{dt} + 2b\frac{db}{dt} = 0 \implies \frac{dh}{dt} = -\frac{b}{h} \frac{db}{dt}}$

$\displaystyle{=-\frac{b}{\sqrt{256 - b^2}} \frac{db}{dt} = -\frac{8}{13.856} \times 1 = -0.588}$

The minus sign indicates that the value of h is actually decreasing. The required answer is 0.588.

b. From the diagram, infer that $16 \sin{\theta} = b$ . When b = 8, then $\theta = \arcsin{0.5} = \ang{30}$ .

Differentiate the above equation w.r.t t

$\displaystyle{16 \cos{\theta} \frac{d\theta}{dt} = \frac{db}{dt} \implies \frac{d\theta}{dt} = \frac{1}{16 \cos{\theta}} = \frac{1}{13.856} = \mathbf{0.0722}}$

c. The area of the triangle is given by $A = 0.5\times h \times b$ . Differentiating w.r.t t,

$\displatstyle{\frac{dA}{dt} = 0.5 b \frac{dh}{dt} + 0.5 h \frac{db}{dt}}$

Plugging in b = 8, h = 13.856, $\frac{dh}{dt} = -0.588$ ,

$\frac{dA}{dt} = -2.352 + 6.928 = \mathbf{4.576 ft^2/sec}$

8 0

4 years ago