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Vlad1618 [11]
3 years ago
15

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

7

Hope this helps! ;)

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C because it’s just it srry if it’s not but I think it is
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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:

It's A

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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:

<em>C.</em> (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]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(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+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

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
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