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2 years ago

Use the distributive property to find the product

Mathematics

2 answers:

dalvyx [7]2 years ago

6 0

Answer:

1st one: -24a-6b-36c

2nd one: 14mn-12m

Step-by-step explanation:

1st one: Multiply '-6' w/ '4a', 'b', and '6c'

2nd one: Multiply '2m' w/ '7n' and '-6'

Simora [160]2 years ago

3 0

The second one is 14mn - 12m and the first is -24a + -6b + -36c

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PLS HELPP! :)<br><br>What is the radius with a circumference of 122 in?

Mumz [18]

Answer:

this is the answer r≈19.42in

because you radius with a circumference of 122in

and that's your answer

Step-by-step explanation:

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3 years ago

£6000 is divided between Adam, Ben and<br> Chris in the ratio 1:3:4. How much does Ben receive?

Gala2k [10]

Answer:

Chris gets £3000

Step-by-step explanation:

A : B : C

1 : 3 : 4 = 8 (add all the parts together)

£6000 ÷ 8 = 750

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2 years ago

Choose the three shapes that have an area of 36 square centimeters

Artemon [7]

Choose the three shapes that have and Area of 36 square centimeters

Step-by-step explanation:

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2 years ago

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