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3 years ago

A portion of the graph of f(x) = −x2 − 2x + 8 is shown. Which of the following describes all solutions for f(x)?

Mathematics

1 answer:

Naily [24]3 years ago

7 0

Answer:

$(x,-x^2-2x+8)$ where −5 ≤ x ≤ 3

Step-by-step explanation:

The given function is $f(x)=-x^2-2x+8$ .

We want to select the option that describes all the solutions to the parabola.

The domain of the parabola is −5 ≤ x ≤ 3.

This means that any x=a on −5 ≤ x ≤ 3 that satisfies (a,f(a)), is a solution.

This can be rewritten as $(a,-a^2-2a+8)$

Therefore for x belonging to −5 ≤ x ≤ 3, all solutions are given by:

$(x,-x^2-2x+8)$ where −5 ≤ x ≤ 3.

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A—15<br> b-30<br> c-36<br> d-72asap

3241004551 [841]

Answer:

C 36

Step-by-step explanation:

1/2(base)(height)

1/2(8)(9)

1/2(72)

=36

8 0

2 years ago

How do you graph y=2/3x-4

Masja [62]

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▹ Answer

You can use a graphing calculator.

▹ Step-by-Step Explanation

Attached is a screenshot.

Hope this helps!

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8 0

3 years ago

Read 2 more answers

Marcus drew a quadrilateral with only 1 pair of parallel sides. Which quadrilateral did Marcus draw? parallelogram rhombus trape

SpyIntel [72]

Answer:

Trapezoid

Step-by-step explanation:

We have been given that Marcus drew a quadrilateral with only 1 pair of parallel sides. We are asked to determine the quadrilateral that Marcus drew.

We know that a parallelogram has two pair of parallel sides.

We also know that rhombus is a parallelogram whose 4 sides are equal is measure and rectangle is a parallelogram with opposite sides equal.

Since parallelogram, rhombus and rectangle has two pair of parallel sides, therefore, Marcus didn't draw any of these.

We know that trapezoid has only one pair of parallel sides, therefore, Marcus drew a trapezoid.

8 0

3 years ago

Shinto is serving pizza. He serves 1/5 of the pizza to one friend and 5/12 to another friend. If he had 9/10 of a pizza to start

never [62]

Answer: 17/60

Step-by-step explanation:

Change all of the fractions to have a common denominator:

LCM of 10, 12, 15 is 60

You get 12/60, 25/60, and 54/60.

54/60 is what he had at the start, and so you can just subtract what he gave away

54/60-12/60=42/60

42/60-25/60=17/60

17 is a prime number so you cannot simplify the fraction

3 0

3 years ago

Read 2 more answers

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

6 0

3 years ago