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victus00 [196]
3 years ago
15

Let ​ f(x)=x2+6x−16 ​. Enter the x-intercepts of the quadratic function in the boxes. and

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
8 0

Answer:

x = - 8, x = 2

Step-by-step explanation:

Given

f(x) = x² + 6x - 16

To find the x- intercepts equate f(x) to zero, that is

x² + 6x - 16 = 0 ← in standard form

(x + 8)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 8 = 0 ⇒ x = - 8

x - 2 = 0 ⇒ x = 2

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Andreas93 [3]

The amount to be paid in rent after 2 years if the rent as of now is $3,000 will be; $3,213.675

The question allows that we choose the amount being paid as rent as of now.

Let the rent paid as of now be; $3,000

In essence; after the first year; the amount increases by 3.5% to become;

  • 1.035 × $3,000

  • = $3,105.

After the second year; we have;

  • 1.035 × $3,105

  • = $3,213.675

Ultimately; the amount to be paid after 2 years will be; $3,213.675.

When given the opportunity to change rent contracts;

  • A situation that will be beneficial would be a 3.5% reduction in rent per year
  • A situation that will not be beneficial would be a 7% increase in rent per year.

Read more;

brainly.com/question/24712879

8 0
2 years ago
A helicopter is hovering 100 meters above the ocean. Directly below it, a submarine is at a depth of 200 meters.The altimeter on
Nitella [24]
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3 0
3 years ago
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Answer:

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6 0
3 years ago
Need help finding intervals where the function is decreasing.
umka21 [38]

f(x)=-5x-5\sin x\implies f'(x)=-5-5\cos x

f is decreasing on some interval (a,b) if f'(x) for all x\in(a,b). Similarly, f is increasing on (a,b) if f'(x)>0 for all x\in(a,b).

On the domain 0\le x\le2\pi, we have

-5-5\cos x=0\implies\cos x=-1\implies x=\pi

as the only critical point. So we need to check the sign of f' on two intervals, (0,\pi) and (\pi,2\pi).

On (0,\pi), pick any test value. Suppose we take x=\dfrac\pi2. Then f'\left(\dfrac\pi2\right)=-5, so f is decreasing on this interval.

On (\pi,2\pi), you'd find that f'>0, so f is increasing on this interval.

8 0
2 years ago
Solve for x: √(32^0 + 2/3)=(0.6)^2-3x​
malfutka [58]

Step-by-step explanation:

<u>Given</u>: √{32⁰ + (2/3)} = (0.6)²⁻³ˣ

<u>Asked</u>: Find the value of x = ?

<u>Solution:</u> Given that √{32⁰ + (2/3)} = (0.6)²⁻³ˣ

⇛√{1 + (2/3)} = (0.6)²⁻³ˣ

⇛√{(1/1) + (2/3) = (0.6)²⁻³ˣ

⇛√{(1*3 + 2*1)/3} = (0.6)²⁻³ˣ

⇛√{(3 + 2)/3} = (0.6)²⁻³ˣ

⇛√(5/3) = (0.6)²⁻³ˣ

Squaring on both sides then

⇛{√(5/3)}² = {(0.6)²⁻³ˣ}²

⇛√(5²/3²) = {0.6}(²⁻³ˣ)²

⇛√{(5*5)/(3*3)} = {0.6}(²⁻³ˣ)²

⇛5/3 = {0.6}(²⁻³ˣ)²

[\mathsf{\because} (aᵐ)ⁿ = aᵐⁿ]

⇛5/3 = (0.6)²*²⁻³ˣ*²

⇛5/3 = (0.6)⁴⁻⁶ˣ

⇛5/3 = (6/10)⁴⁻⁶ˣ

⇛5/3 = {(6÷2)/(10÷2)})⁴⁻⁶ˣ

⇛5/3 = (3/5)⁴⁻⁶ˣ

⇛(3/5))⁻¹ = (3/5)⁴⁻⁶ˣ

[\mathsf{\because} a⁻ⁿ = 1/aⁿ]

Base are the same, so the exponents must be equal.

\mathsf{\therefore} -1 = 4 - 6x

Shift the number 4 from RHS to LHS, changing it's sign.

⇛-1 - 4 = -6x

⇛-5 = -6x

⇛x = {(-5)/(-6)}

\mathsf{\therefore} x = 5/6

<u>Answer</u><u>:</u> Hence, the value of x for the given problem is 5/6.

<u>also </u><u>read </u><u>similar </u><u>questions</u><u>:</u> (1/3)^-2 + (1/4)^-2 = (125)^x Find the value of x? brainly.com/question/25746379?referrer

(3^-2 × 4^-2)^-3 = 12^x In the given expression find the value of x. brainly.com/question/25746273?referrer

4 0
2 years ago
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