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3 years ago

12m^8 × (2m^2)^3 ÷ 4m^5

Mathematics

2 answers:

Dmitry_Shevchenko [17]3 years ago

6 0

Answer:

12m8(2m2)3

=24m19

Step-by-step explanation:

Roman55 [17]3 years ago

3 0

Answer: =24m^19

Step-by-step explanation:

12m^8 × (2m^2)^3 ÷ 4m^5

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2 years ago

Consider this equation. 7.8+2(0.75m+0.4)=-6.4m+4(0.5m-0.8) Mia solved the equation and determined that m=2. Is she correct?

IgorC [24]

Answer:

B) She is incorrect because when substituting 2 for m the result was a false statement.

Step-by-step explanation:

Let's solve it for ourselves and see if Mia is correct:

7.8+2(0.75m+0.4)=-6.4m+4(0.5m-0.8)

7.8+1.5m+0.8=-6.4m+2m-3.2

1.5m+8.6=-4.4m-3.2

8.6=-5.9m-3.2

11.8=-5.9m

-2=m

m=-2

Since the answer is actually m=-2, Mia's guess is incorrect because plugging in m=2 does not make both sides equal. Therefore B is the correct option.

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2 years ago

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2 years ago

An arrow is shot vertically upward from a platform 20 ft high at a rate of 179ft / s * e When will the arrow hit the ground ? Us

MissTica

The height attained by the arrow is modelled as,

$h(t)=-16t^2-v_{\circ}t+h_{\circ}$

Given that the initial height is 20 ft and the initial velocity is 179 ft per second,

$\begin{gathered} v_{\circ}=179 \\ h_{\circ}=20 \end{gathered}$

Substitute the values,

$\begin{gathered} h(t)=-16t^2-(179)t+(20) \\ h(t)=-16t^2-179t+20 \end{gathered}$

When the arrow hits the ground its height will become zero,

$\begin{gathered} h(t)=0 \\ -16t^2-179t+20=0 \end{gathered}$

Applying the quadratic formula,

$\begin{gathered} t=\frac{-(-179)\pm\sqrt[]{(-179)^2-4(-16)(20)}}{2(-16)} \\ t=\frac{179\pm\sqrt[]{33321}}{-32} \\ t=\frac{179\pm182.54}{-32} \\ t=\frac{179+182.54}{-32},t=\frac{179-182.54}{-32} \\ t=-11.298,t=0.1106 \end{gathered}$

Since time cannot be negative, we have to neglect the negative value.

Thus, it can be concluded that the arrow will hit the ground after 0.1106 seconds approximately.

7 0

8 months ago

12m^8 × (2m^2)^3 ÷ 4m^5​

12m^8 × (2m^2)^3 ÷ 4m^5