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4 years ago

What is the value of (-6)3? A. -18 B. 18 C. -216 D. 216

Mathematics

2 answers:

gogolik [260]4 years ago

5 0

A. The answer is negative eighteen

kiruha [24]4 years ago

3 0

B is the answer to your equation as long as you're not meaning $-6^{3}$ . Then your answer will be C.

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16. The temperature of an oven at 0°C has an

butalik [34]

Answer:

300K

Step-by-step explanation:

T = t + 273 °C

T is absolute temperature

t = given temperature (27°C in this case)

Hence T = 27 °C + 273 °C (standard temperature)

T = 300K (since absolute temperature is measured in K from the question's preamble)

6 0

3 years ago

Solve -2x ≤ -4. Graph the solution

horsena [70]

Https://us-static.z-dn.net/files/d2a/4dd214e6e402aa7e840049f27983c856.jpeg

3 0

3 years ago

You’re right square format has a base area of 200 inches

Dovator [93]

Answer:

1200 in³

Step-by-step explanation:

V= l×w×h

they gave you l×w which is 200.

200×6= 1200

6 0

3 years ago

Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

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Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$

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Part III

$\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\$

7 0

3 years ago

Q1.Simplify:

solniwko [45]

$\textbf{(i)}\\\\\{1^3 +2^3 \} \times \left(\dfrac 13 \right)^2\\\\=(1+8) \left(\dfrac 19 \right)\\\\=9\left(\dfrac 19 \right)\\\\=1\\\\$

$\textbf{(ii)}\\\\\{5^{-1}\times 4^{-1}\}^2\\\\=\left(5^{-1} \right)^2 \times \left(4^{-1} \right)^2~~~~~~~~~~~;[(ab)^m = a^mb^m]\\\\=5^{-2}\times 4^{-2}~~~~~~~~~~~~~~~~~~~~;[(a^m)^n = a^{mn}]\\\\=\dfrac 1{5^2} \times \dfrac 1{4^2}~~~~~~~~~~~~~~~~~~~~~~;\left[a^{-m} = \dfrac 1{a^m},~ a\neq 0 \right]\\\\=\dfrac{1}{400}\\\\=0.0025\\\\$

$\textbf{(iii)}\\\\\left\{ \left(\dfrac 13 \right)^{-2} \times \left( \dfrac 12 \right)^{-2}\right\}\div \left(\dfrac 14 \right)^{-3}\\\\\\=\left( \dfrac 13 \times \dfrac 12 \right)^{-2} \div\left(4^{-1}\right)^{-3}\\\\\\=\left(\dfrac 16 \right)^{-2} \div 4^3\\\\\\=\left(6^{-1} \right)^{-2}\div 4^3\\\\\\=6^2 \div 4^3\\\\\\=36\div 64\\\\\\=\dfrac{9}{16}\\\\\\=0.5625$

6 0

2 years ago

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