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4 years ago

Complete the square x^2+6x-16=0

2 answers:

7 0

X=2 and x= -8
x^2+6x-16=0
1) Move the constant (-16) to the right of the equal sign: x^2+6x = 16 (becomes positive)
2) Make the left side a perfect trinomial by doing half of 6 (3) then squaring it (9). Add 9 to the 16: x^2+6x+9= 16+9.
3) Factor the left side, then find the sum of the right side: (x+3)^2 = 25
4) Take the square root of both sides: x+3 =(+/-) 5 (remember it's both positive & negative).
The equation will look like x+3= - 5 and x+3=5

5) solve the two equations: x= -8 and x=2

When you put the x-values back into the problem, it should come out 0.

Hope this helps

strojnjashka [21]4 years ago

6 0

Let's solve your equation step-by-step:
x^2+6x-16=0
Factor the left side of the equation:(x-2)(x+8)=0
Set factors equal to 0: x-2=0 or x+8=0, x=2 or x=-8
Your answer is x=2 or x=-8, doesn't matter which, both are correct.
Hope this helps!:)

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Its a math test again-

Step2247 [10]

Answer:

A. t > 2.5

Step-by-step explanation:

If she wants to achieve <em>more </em>than 590, we already know the sign would be greater than (>).

All we have to do is divide 590/236, which gives us 2.5.

Therefore, she will have to ride more than 2.5 hours to reach her goal, t > 2.5

Hope this helps :)

Good luck on your test.

3 0

3 years ago

From exterior point L of circle O, tangent segments LP and LQ are drawn such that the measure of angle PLQ is 60 degrees. If a r

IrinaK [193]

Wow ! There's so much extra mush here that the likelihood of being
distracted and led astray is almost unavoidable.

The circle ' O ' is roughly 98.17% (π/3.2) useless to us. The only reason
we need it at all is in order to recall that the tangent to a circle is
perpendicular to the radius drawn to the tangent point. And now
we can discard Circle - ' O ' .
Just keep the point at its center, and call it point - O .

-- The segments LP, LQ, and LO, along with the radii OP and OQ, form
two right triangles, reposing romantically hypotenuse-to-hypotenuse.
The length of segment LO ... their common hypotenuse ... is the answer
to the question.

-- Angle PLQ is 60 degrees. The common hypotenuse is its bisector.
So the acute angle of each triangle at point ' L ' is 30 degrees, and the
acute angle of each triangle at point ' O ' is 60 degrees.

-- The leg of each triangle opposite the 30-degree angle is a radius
of the discarded circle, and measures 6 .

-- In every 30-60 right triangle, the length of the side opposite the hypotenuse
is one-half the length of the hypotenuse.

-- So the length of the hypotenuse (segment LO) is <em>12 </em>.

4 0

3 years ago

Anu is a photographer and was recently hired to photograph a wedding in Hawaii. While there, Anu had some time to

andriy [413]

ANSWER: 480 out of 1200 were taken at the Honolulu Zoo.

5 0

2 years ago

If the 2nd and 5th terms of a

blagie [28]

Answer:

Step-by-step explanation:

The n th term of a GP is

$a_{n}$ = a $r^{n-1}$

where a is the first term and r the common ratio

Given a₂ = 6 and a₅ = 48, then

ar = 6 → (1)

a $r^{4}$ = 48 → (2)

Divide (2) by (1)

$\frac{ar^4}{ar}$ = $\frac{48}{6}$ , that is

r³ = 8 ( take the cube root of both sides )

r = $\sqrt[3]{8}$ = 2

Substitute r = 2 into (1)

2a = 6 ( divide both sides by 2 )

a = 3

Thus

3, 6, 12, 24 ← are the first 4 terms

3 + 6 + 12 + 24 = 45 ← sum of first 4 terms

8 0

3 years ago

Find the angle between the given vectors to the nearest tenth of a degree.

saul85 [17]

Answer:

Step-by-step explanation:

Given

u = <6, -1>

u = 6i-j

and

v=<7,-4>

v=7i-4j

The formula for angle is:

Let x be the angle

$cos\ x = \frac{u.v}{||u||.||v||}$

where ||u|| is the length and u.v is the dot product or scalar product of both vectors

So,

$||u|| = \sqrt{(6)^2+(-1)^2}\\ = \sqrt{36+1}\\ = \sqrt{37}\\ ||v||=\sqrt{(7)^2+(-4)^2}\\ = \sqrt{49+16}\\ = \sqrt{65}\\$

$u.v = u_1u_2+v_1v_2\\= (6)(7)+(-1)(-4)\\=42+4\\=46$

$cos\ x=\frac{46}{\sqrt{37}\sqrt{65}} \\= \frac{46}{\sqrt{2405} }\\Can\ also\ be\ written\ as:\\= \frac{46}{\sqrt{2405} } * \frac{\sqrt{2405} }{\sqrt{2405}} \\=\frac{46\sqrt{2405} }{2405}$

The calculated angle will be in radians. To find the angle in degrees:

$x = \frac{180}{\pi} cos^{-1} (\frac{46\sqrt{2405} }{2405})\\x = 20.282\\x= 20.3\\$

Hence Option A is correct ..

6 0

4 years ago