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4 years ago

Steps of sloving system of equation steps

Mathematics

1 answer:

MariettaO [177]4 years ago

6 0

Answer:

Step-by-step explanation:

Solve 1 equation for 1 variable. ( Put in y = or x = form)

Substitute this expression into the other equation and solve for the missing variable.

Substitute your answer into the first equation and solve.

Check the solution.

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How to solve logarithmic equations as such

Serga [27]

$\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5$

$\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}$

5 0

3 years ago

A company packages its milk powder in cylindrical container whose base has a diameter of 28 cm and

yKpoI14uk [10]

Answer:

896pi or 2814.9cm^2

Step-by-step explanation:

S=h*r*2pi=(40-2*4)*28/2*2pi=32*14*2pi=896pi=2814.9cm^2

6 0

3 years ago

Please help! Sketch the graph of each line.

madam [21]

Answer:

1) (-3,0) (0,-3)

2) (-2,0)

3) (5,0) (0,-1)

4) (12,0) (0,3)

5) (-16,0) (0,4)

6) (-0.8333,0) (0,5)

4 0

3 years ago

3. A balancing balloon toy is in the shape of a hemisphere (half-sphere) attached to the base of a cone. If the toy is 4ft tall

Katen [24]

Answer:

The volume of the toy is $V=5.23\ ft^3$

Step-by-step explanation:

step 1

Find the volume of the hemisphere

The volume of the hemisphere is given by the formula

$V=\frac{2}{3}\pi r^{3}$

In this problem, the wide of the toy is equal to the diameter of the hemisphere

$D=2\ ft$

$r=2/2=1\ ft$ ----> the radius is half the diameter

substitute

$V=\frac{2}{3} \pi (1)^{3}=\frac{2}{3} \pi\ ft^3$

step 2

Find the volume of the cone

The volume of the cone is given by

$V=\frac{1}{3}\pi r^{2}h$

we know that

The radius of the cone is the same that the radius of the hemisphere

$r=1\ ft$

The height of the cone is equal to subtract the radius of the hemisphere from the height of the toy

$h=4-1=3\ ft$

substitute the given values

$V=\frac{1}{3}\pi (1)^{2}(3)=\pi\ ft^3$

step 3

Find the volume of the toy

we know that

The volume of the toy, is equal to the volume of the cone plus the volume of the hemisphere.

$V=(\frac{2}{3} \pi+\pi)\ ft^3$

$V=(\frac{5}{3}\pi)\ ft^3$

assume

$\pi=3.14$

$V=\frac{5}{3}(3.14)=5.23\ ft^3$

5 0

4 years ago

Tom was using wire of the following thicknesses .33 mm, .275 mm, .25 mm, and .3 mm for some electrical work. Order the wire from

RideAnS [48]

$\bf \begin{array}{ll|ll}
&padded&sorted&\stackrel{unpadded}{sorted}\\
----&----&----&----\\
.33&0.330&330&.33\\
.275&0.275&300&.3\\
.25&0.250&275&.275\\
.3&0.300&250&.25\\
\end{array}$

7 0

3 years ago

Read 2 more answers