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3 years ago

If a number is negative, then its opposite will be:

Mathematics

1 answer:

Arturiano [62]3 years ago

7 0

Answer:

the opposite of a negative number is positive.

Step-by-step explanation:

Negative numbers are numbers that are less than 0. The opposite of a positive number is negative, and the opposite of a negative number is positive. Since the opposite of 0 is 0 (which is neither positive nor negative), then 0 = 0.

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G(x)=3x-4 find (gog) (x)

Anestetic [448]

Answer:

gog(x)=9x-16

Step-by-step explanation:

We are given g(x)=3x-4 and are asked to find (gog) (x)

gog(x)=g(g(x))

g(x)=3x-4

Hence

g(g(x))=g(3x-4)

=3(3x-4)-4

=9x-12-4

=9x-16

Hence

gog(x)=9x-16

5 0

3 years ago

Clarissa helps her mom put the 200-Newton lawn mower in the back of her mom’s truck. They lift the mower up from the ground to t

Anarel [89]

Answer:

220 N * m or 220 J

Step-by-step explanation:

We have that the work formula would be the applied force multiplied by the distance traveled, that is:

W = F * d

In this case the F is 200 N and the distance is 1.1 meters, therefore we replace:

W = 200 * 1.1

W = 220 N * m = 220 J

Therefore the work carried out by Clarissa and her mom is 200 joules.

4 0

3 years ago

Help me please!!!!!!!!!!!!!

Virty [35]

Hiiiiiii

The answer is 1 1/9

5 0

3 years ago

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Write an expression for the sequence of operations described below. divide 10 by the sum of n and m Do not simplify any part of

lina2011 [118]

Answer:

(m+n)/10

Step-by-step explanation:

divide 10 by the sum of n and m

(m+n)/10

3 0

3 years ago

Read 2 more answers

Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−

aleksandrvk [35]

The tangent to $C$ through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces $S$ and $T$ at that point.

Let $f(x,y,z)=x^2+2y^3+3z^4$ . Then $S$ is the level curve $f(x,y,z)=6$ . Recall that the gradient vector is perpendicular to level curves; we have

$\nabla f(x,y,z)=(2x,6y,12z^2)$

so that the gradient of $f$ at (1, 1, 1) is

$\nabla f(1,1,1)=(2,6,12)$

For the surface $T$ , we have

$\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0$

so that $\vec r(1,0)=(1,1,1)$ . We can obtain a vector normal to $T$ by taking the cross product of the partial derivatives of $\vec r(u,v)$ , and evaluating that product for $u=1,v=0$ :

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)$

$\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)$

Now take the cross product of the two normal vectors to $S$ and $T$ :

$(2,6,12)\times(1,-1,2)=(24,8,-8)$

The direction of vector (24, 8, -8) is the direction of the tangent line to $C$ at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by $t\in\Bbb R$ . Then adding (1, 1, 1) shifts this line to the point of tangency on $C$ . So the tangent line has equation

$\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)$

7 0

3 years ago

If a number is negative, then its opposite will be:​

If a number is negative, then its opposite will be: