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2 years ago

Can you tell me what is 26-4s=9s

2 answers:

8 0

Remember that you can do ANYTHING to an equation as long as you do it to both sides

26-4s=9s
addd 4s to both sides
26+4s-4s=9s+4s
26+0=13s
26=13s
divide both sideis by 13
26/13=(13s)/13
2=(13/13)s
2=1s
2=s
s=2

wariber [46]2 years ago

6 0

26 - 4s = 9s
26 = 9s + 4s
26 = 13s
13s = 26
s = 26/13
s = 2

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the area is 50.24

Step-by-step explanation:

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2 years ago

Find the first three terms in the expansion , in ascending power of x , of (2+x)^6 and obtain the coefficient of x^2 in the expa

Nataly_w [17]

Answer:

The first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

coefficient of $x^{2}$ in the expansion of $(2+x - x^{2})^{6}$ = (240 - 192) = 48

Step-by-step explanation:

$(2+x)^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times x^{k} \times 2^{6 - k})$

= $6_{C_{0}} \times x^{0} \times 2^{6} + 6_{C_{1}} \times x^{1} \times 2^{5} + 6_{C_{2}} \times x^{2} \times 2^{4}$ + terms involving higher powers of x

= $64 + 192 \times x^{1} + 240 \times x^{2}$ + terms involving higher powers of x

so, the first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

Again,

$(2+x - x^{2})^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times (2 + x)^{k} \times (-x^{2})^{6 - k})$

Now, by inspection,

the term $x^{2}$ comes from k =5 and k = 6

for k = 5, the coefficient of $x^{2}$ is , $(-32) \times 6$ = -192

for k = 6 , the coefficient of $x^{2}$ is, $6_{C_{2}} \times 2^{4}$ = 240

so, coefficient of $x^{2}$ in the final expression = (240 - 192) = 48

3 0

2 years ago

enelope determined the solutions of the quadratic function by completing the square. f(x) = 4x2 + 8x + 1 –1 = 4x2 + 8x –1 = 4(x2

Marat540 [252]

The correct question is
Penelope determined the solutions of the quadratic function by completing the square.
f(x) = 4x² + 8x + 1
–1 = 4x² + 8x
–1 = 4(x² + 2x)
–1 + 1 = 4(x² + 2x + 1)
0 = 4(x + 2)²
0 = (x + 2)²
0 = x + 2
–2 = x
What error did Penelope make in her work?

we have that
f(x) = 4x² + 8x + 1
to find the solutions of the quadratic function
let
f(x)=0
4x² + 8x + 1=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(4x² + 8x)=-1

Factor the leading coefficient

4*(x² + 2x)=-1

Complete the square Remember to balance the equation by adding the same constants to each side.

4*(x² + 2x+1)=-1+4 --------> ( added 4 to both sides)

Rewrite as perfect squares

4*(x+1)²=3

(x+1)²=3/4--------> (+/-)[x+1]=√3/2

(+)[x+1]=√3/2---> x1=(√3/2)-1----> x1=(√3-2)/2

(-)[x+1]=√3/2----> x2=(-2-√3)/2

therefore

the answer is

Penelope should have added 4 to both sides instead of adding 1.

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