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3 years ago

Substitute the value of x back into the equation x + 3 -

Mathematics

1 answer:

statuscvo [17]3 years ago

5 0

2x = -1 + 1x + 5

-1x -1x

x = -1 + 5

x = 4

x + 3

4 + 3

Hope this helps! ;)

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The graph shows the number of acres that mr.Henry has used on his farm for baking hay over several years. Using a line of best f

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3 years ago

Como simplifica essa expressão: x+6y-7z+2-8x-12y-10z-20:

n200080 [17]

Answer:

-17 z - 6 y - 7 x + -18

Step-by-step explanation:

Simplify the following:

x + 6 y - 7 z + 2 - 8 x - 12 y - 10 z - 20

Hint: | Group like terms in x + 6 y - 7 z + 2 - 8 x - 12 y - 10 z - 20.

Grouping like terms, x + 6 y - 7 z + 2 - 8 x - 12 y - 10 z - 20 = (-7 z - 10 z) + (6 y - 12 y) + (x - 8 x) + (2 - 20):

(-7 z - 10 z) + (6 y - 12 y) + (x - 8 x) + (2 - 20)

Hint: | Combine like terms in -7 z - 10 z.

-7 z - 10 z = -17 z:

-17 z + (6 y - 12 y) + (x - 8 x) + (2 - 20)

Hint: | Combine like terms in 6 y - 12 y.

6 y - 12 y = -6 y:

-17 z + -6 y + (x - 8 x) + (2 - 20)

Hint: | Combine like terms in x - 8 x.

x - 8 x = -7 x:

-17 z - 6 y + -7 x + (2 - 20)

Hint: | Evaluate 2 - 20.

2 - 20 = -18:

Answer: -17 z - 6 y - 7 x + -18

3 0

3 years ago

What is the name of the shape depicted in the graph below?

Ahat [919]

Answer:

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Step-by-step explanation:

6 0

3 years ago

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For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago

a rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y=121-x^2. what are the dimension

insens350 [35]

Answer:

the length is $\dfrac{22}{\sqrt{3}}$ and the width is $121-\left(\dfrac{11}{\sqrt{3}}\right)^2=121-\dfrac{121}{3}=\dfrac{242}{3}.$

Step-by-step explanation:

Let points A and B be placed on the x-axis. Their coordinates are $A(x_0,0)\ (x_0>0)$ and $B(-x_0,0)$ (because of parabola symmetry). Two other vertices lie on the parabola, then $C(-x_0,121-x_0^2)$ and $D(x_0,121-x_0^2).$ The length of the side AB is $2x_0$ and the length of the side AD is $121-x_0^2.$ Thus, the area of the rectangle ABCD is

$A=2x_0\cdot (121-x_0^2)=242x_0-2x_0^3.$

Find the derivative A':

$A'=242-2\cdot 3x_0^2=242-6x_0^2.$

Equate A' to 0:

$242-6x_0^2=0,\\ \\x_0^2=\dfrac{121}{3},\\ \\x_0=\dfrac{11}{\sqrt{3}}.$

The maximum area of the rectangle is

$A_{max}=242\cdot \dfrac{11}{\sqrt{3}}-2\cdot \left(\dfrac{11}{\sqrt{3}}\right)^3=\dfrac{2662}{\sqrt{3}}-\dfrac{2662}{3\sqrt{3}}=\dfrac{5324}{3\sqrt{3}}\ un^2.$

The dimensions of the rectangle are:

the length is $\dfrac{22}{\sqrt{3}}\ un.$ and the width is $121-\left(\dfrac{11}{\sqrt{3}}\right)^2=121-\dfrac{121}{3}=\dfrac{242}{3}\ un.$

6 0

4 years ago

Read 2 more answers