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3 years ago

Given the function g(x)=x^2-2 find the range for the domain (-2,-1,1,3)

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

5 0

Answer:

(-2,2)

(-1,-1)

(1,-1)

(3,7)

If you graph it with a graphing calc. and pinpoint the X's(domains) then you can see the Y's(ranges)

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Find the solution set 7x^2-9x-10=0

vitfil [10]

Answer:

x = -5/7, 2

Step-by-step explanation:

Step 1: Factor

(5x + 7)(x - 2) = 0

Step 2: Find <em>x </em>roots

5x + 7 = 0

x = -5/7

x - 2 = 0

x = 2

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3 years ago

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A football team gained 5 yards and then last 7 yards. What real number represents the teams position relative to its original po

lana66690 [7]

Answer:

$\boxed{-2}}$

Step-by-step explanation:

Think of this situation as a number line in which the team starts at 0 yd.

Gaining yards is the positive direction and losing yards is negative.

Start at 0 yd and go 5 yd to the right. You are at +5.

Then go 7 yd left. You end up at -2.

+5 – 7 = -2

$\text{The real number that represents the team's position relative to its original position is $\boxed{\mathbf{-2}}$}$

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3 years ago

102° find the value of x and x

Whitepunk [10]

Answer:

17 3.14

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Step-by-step explanation:

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3 years ago

Describe how adding and subtracting mixed numbers can help you with recipes.

andrey2020 [161]

Because there's times when u need to double for instance a cup so u have to add one morecup

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3 years ago

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evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−yj+zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.

meriva

We have

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt$

and

$\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k$

$\vec r(t) = \sin(t)\,\vec\imath + \cos(t)\,\vec\jmath + t\,\vec k \implies \dfrac{d\vec r}{dt} = \cos(t) \, \vec\imath - \sin(t) \, \vec\jmath + \vec k$

so the line integral is equilvalent to

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}$

7 0

2 years ago