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bulgar [2K]
4 years ago
14

Plz help me with this math and also explain

Mathematics
1 answer:
jasenka [17]3 years ago
8 0

Step-by-step explanation:

<h2>[1]</h2>

  • SI = $250
  • Rate (R) = 12\sf \dfrac{1}{2} %
  • Time (t) = 4 years

\longrightarrow \tt { SI = \dfrac{PRT}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 12\cfrac{1}{2} \times 4}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times \cfrac{25}{2} \times 4}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 25 \times 2}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 50}{100} } \\

\longrightarrow \tt { 250 \times 100 = P \times 50} \\

\longrightarrow \tt { 25000 = P \times 50} \\

\longrightarrow \tt { \dfrac{25000}{50} = P } \\

\longrightarrow \underline{\boxed{ \green{ \tt { \$ \; 500 = P }}}} \\

Therefore principal is $500.

<h2>__________________</h2>

<h2>[2]</h2>

  • 2/7 of the balls are red.
  • 3/5 of the balls are blue.
  • Rest are yellow.
  • Number of yellow balls = 36

Let the total number of balls be x.

→ Red balls + Blue balls + Yellow balls = Total number of balls

\longrightarrow \tt{ \dfrac{2}{7}x + \dfrac{3}{5}x + 36 = x} \\

\longrightarrow \tt{ \dfrac{10x + 21x + 1260}{35} = x} \\

\longrightarrow \tt{ \dfrac{31x + 1260}{35} = x} \\

\longrightarrow \tt{ 31x + 1260= 35x} \\

\longrightarrow \tt{ 1260= 35x-31x} \\

\longrightarrow \tt{ 1260= 4x} \\

\longrightarrow \tt{ \dfrac{1260 }{4}= x} \\

\longrightarrow \underline{\boxed{  \tt { 315 = x }}} \\

Total number of balls is 315.

A/Q,

3/5 of the balls are blue.

\longrightarrow \tt{ Balls_{(Blue)} =\dfrac{3 }{5}x} \\

\longrightarrow \tt{ Balls_{(Blue)} =\dfrac{3 }{5}(315)} \\

\longrightarrow \tt{ Balls_{(Blue)} = 3(63)} \\

\longrightarrow \underline{\boxed{ \green {\tt { Balls_{(Blue)} = 189 }}}} \\

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4m exponent 2 + 5m<br><br> help please
Gnesinka [82]

\huge \boxed{\mathbb{QUESTION} \downarrow}

  • 4m exponent 2 + 5m.

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

By.. 4m exponent 2, I'm going to assume that you meant this ⇨ 4m². If so, then here's how you should solve your question. I've included 2 ways of solving the question. You can choose your method. I'd suggest the 2nd one since it's a whole lot easier than the first method.

<h3><u>Method </u><u>1</u><u> </u><u>:</u><u>-</u></h3>

\tt \: 4m  ^ { 2  }  +5m \\

Quadratic polynomial can be factored using the transformation \tt\: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where \tt\:x_{1} and x_{2} are the solutions of the quadratic equation ax²+bx+c=0.

\tt \: 4m^{2}+5m=0

All equations of the form ax²+bx+c=0 can be solved using the quadratic formula: \tt\frac{-b±\sqrt{b^{2}-4ac}}{2a}\\. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

\tt \: m=\frac{-5±\sqrt{5^{2}}}{2\times 4}  \\

Take the square root of 5².

\tt \: m=\frac{-5±5}{2\times 4}  \\

Multiply 2 times 4.

\tt \: m=\frac{-5±5}{8}  \\

Now solve the equation m=\tt\frac{-5±5}{8} when ± is plus. Add -5 to 5.

\tt \: m=\frac{0}{8}  \\  \\  \tt \: m = 0

Now solve the equation m=\tt\frac{-5±5}{8} when ± is minus. Subtract 5 from -5.

\tt \: m=\frac{-10}{8}  \\

Reduce the fraction -10/8 to its lowest terms by extracting and cancelling out 2.

Factor the original expression using \tt\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for \tt\:x_{1} \:and \:-\frac{5}{4} \:for\: x_{2}.

\tt \: 4m^{2}+5m=4m\left(m-\left(-\frac{5}{4}\right)\right)  \\

Simplify all of the expressions of the form \tt\:p-\left(-q\right) to p+q.

\tt \: 4m^{2}+5m=4m\left(m+\frac{5}{4}\right)  \\

Add 5/4 to m by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

\tt \: 4m^{2}+5m=4m\times \left(\frac{4m+5}{4}\right)  \\

Cancel out 4, the greatest common factor in 4 and 4.

\tt \: 4m^{2}+5m=   \boxed{\boxed{\bf \: m\left(4m+5\right) }}

<em><u>Don't</u></em><em><u> </u></em><em><u>worry.</u></em><em><u> </u></em><em><u>If </u></em><em><u>this </u></em><em><u>process</u></em><em><u> </u></em><em><u>seems </u></em><em><u>long </u></em><em><u>&</u></em><em><u> </u></em><em><u>wears </u></em><em><u>you</u></em><em><u> </u></em><em><u>out </u></em><em><u>or </u></em><em><u>if </u></em><em><u>you </u></em><em><u>haven't</u></em><em><u> </u></em><em><u>learned</u></em><em><u> </u></em><em><u>the </u></em><em><u>biquadratic</u></em><em><u> formula</u></em><em><u> </u></em><em><u>yet,</u></em><em><u> </u></em><em><u>you </u></em><em><u>can </u></em><em><u>just </u></em><em><u>use </u></em><em><u>the </u></em><em><u>simple </u></em><em><u>method</u></em><em><u> of</u></em><em><u> </u></em><em><u>factoring </u></em><em><u>out </u></em><em><u>the </u></em><em><u>common</u></em><em><u> </u></em><em><u>term </u></em><em><u>(</u></em><em><u>m)</u></em><em><u>.</u></em><em><u> </u></em><em><u>Here's</u></em><em><u> how</u></em><em><u> you</u></em><em><u> </u></em><em><u>do </u></em><em><u>that </u></em>\downarrow

<h3><u>Method</u><u> </u><u>2</u><u> </u><u>:</u><u>-</u></h3>

\tt \: 4m  ^ { 2  }  +5m

Factor out m.

=  \boxed{ \boxed{ \bf \: m\left(4m+5\right) }}

See, the second method is easier. But, if your question comes for a lot of marks then you might prefer using the first method.

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Answer:

17,148.59 in³

Step-by-step explanation:

4/3 pi r^3

4/3(3.14)(16)^3

17148.58667

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Step-by-step explanation:

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