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3 years ago

15

Rewrite by completing the square 2x^2+7x+6=0

1 answer:

Kitty [74]3 years ago

6 0

Answer:

$2x^2+7x+6=0\\\\2(x^2+\frac72x+3)=0\\\\2[x^2+2\cdot x\cdot\frac74+(\frac74)^2-(\frac74)^2+3]=0\\\\2[x^2+2\cdot x\cdot\frac74+(\frac74)^2]-2(\frac74)^2+6=0 \\\\2(x+\frac74)^2-2\cdot\frac{49}{16}+6=0\\\\2(x+\frac74)^2-\frac{49}8+\frac{48}8=0\\\\2(x+\frac74)^2-\frac18=0$

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Help me with this question yall

lana [24]

<u>Solution:</u><u>-</u>

= 6⁴/6²

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3 years ago

If f(x) = 2x^2+3, find f(a+2)

zzz [600]

Answer:

2a²+8a+7

Step-by-step explanation:

6 0

3 years ago

Solve the system of linear equations using linear combination.

snow_lady [41]

Answer:

a = 1, b = 7

Step-by-step explanation:

Given the 2 equations

3a + 6b = 45 → (1)

2a - 2b = - 12 → (2)

Multiply (2) by 3 and add to (1) to eliminate the term in b

6a - 6b = - 36 → (3)

Add (1) and (3) term by term

(3a + 6a) + (6b - 6b) = (45 - 36)

9a = 9 ( divide both sides by 9 )

a = 1

Substitute a = 1 in either (1) or (2) and solve for b

(1) → 3 + 6b = 45 ( subtract 3 from both sides )

6b = 42 ( divide both sides by 6 )

b = 7

8 0

3 years ago

how do u know what is the next number in pi I know it never stops but how do u know which number is next.

Free_Kalibri [48]

Answer:

There is no "next number"

I think.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Write the vector v in the form aiplusbj, given its magnitude left norm Bold v right norm and the angle alpha it makes with the

photoshop1234 [79]

Answer:

$6\sqrt{3}i-6j$

Step-by-step explanation:

Given:

The magnitude, v=12

The angle $\alpha$ it makes with the positive x-axis = $150^0$

Now for a vector (x,y)

$\alpha =tan^{-1}(\frac{y}{x} )$ ; and

$v=\sqrt{x^2+y^2}$

Therefore:

$150 =tan^{-1}(\frac{y}{x} )\\\frac{y}{x} =tan 150\\\frac{y}{x}=-\frac{\sqrt{3} }{3}$

$y=-\frac{x \sqrt{3}}{3}$

Similarly:

$v=\sqrt{x^2+y^2}\\12=\sqrt{x^2+y^2}\\12^2=x^2+y^2\\144=x^2+y^2$

$144=x^2+(-\frac{x \sqrt{3}}{3} )^2\\144=x^2+\frac{3x^2}{9} \\144=\frac{4x^2}{3}\\4x^2=144 X 3=432\\x^2=108\\x=6\sqrt{3}$

Recall that:

$y=-\frac{x \sqrt{3}}{3}\\y=-\frac{6\sqrt{3} \sqrt{3}}{3}=-6$

Therefore: $(x,y)=(6\sqrt{3}, -6)$

The vector in the form ai+bj is therefore:

$6\sqrt{3}i-6j$

6 0

3 years ago