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3 years ago

Simplify 3a2 x 5b3 x 4a x 2c

Mathematics

1 answer:

sergey [27]3 years ago

6 0

Here you go!! Hope this helps

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Partition a line segment in the given ratio.

mestny [16]

Check the picture below.

$\textit{internal division of a line segment using ratios} \\\\\\ T(-5,9)\qquad U(9,2)\qquad \qquad \stackrel{\textit{ratio from T to U}}{5:2} \\\\\\ \cfrac{T\underline{V}}{\underline{V} U} = \cfrac{5}{2}\implies \cfrac{T}{U} = \cfrac{5}{2}\implies 2T=5U\implies 2(-5,9)=5(9,2)$

$(\stackrel{x}{-10}~~,~~ \stackrel{y}{18})=(\stackrel{x}{45}~~,~~ \stackrel{y}{10}) \implies V=\underset{\textit{sum of the ratios}}{\left( \cfrac{\stackrel{\textit{sum of x's}}{-10 +45}}{5+2}~~,~~\cfrac{\stackrel{\textit{sum of y's}}{18 +10}}{5+2} \right)} \\\\\\ V=\left( \cfrac{ 35 }{ 7 }~~,~~\cfrac{ 28}{ 7 } \right)\implies {\Large \begin{array}{llll} V=(5~~,~~4) \end{array}}$

4 0

1 year ago

Expand the expression: 4(1.8h - 2)

Sophie [7]

Answer:

7.2h-2

Step-by-step explanation:

Multiply

1.8

7.2

−

7 0

3 years ago

This is what I meant on my last question. sorry

Zigmanuir [339]

Answer:

its C

Step-by-step explanation:

4 0

3 years ago

In 2003, the price of a certain automobile was approximately $32,000 with a depreciation of $1,740 per year. After how many year

IRINA_888 [86]

Answer:

The correct answer is 5 years i.e. 2008.

Step-by-step explanation:

Price of the automobile in 2003 is $32000.

Depreciation per year is given by $1740.

Therefore let the car value is depreciated for t number of years.

Value depreciated for t years is given by $ (1740t).

The final value of the car after t years is given to be $23300.

Thus the equation is given by 32000 - 1740t = 23300.

⇒ 1740t = 32000 - 23300

⇒ 1740t = 8700

⇒ t = 5

Thus after 5 years the value of the car is $23300.

Thus in 2008 the depreciated price of the car would be $23300.

7 0

4 years ago

I need help with this guys

natka813 [3]

Answer:

9/4 x 16/5

=36/5

Step-by-step explanation:

7 0

3 years ago