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4 years ago

What is on left side of equation if f=11

Mathematics

1 answer:

ad-work [718]4 years ago

8 0

Answer: f

Step-by-step explanation:

11 is on RHS or the right hand side

therefore the variable in LHS or left hand side is f

hope it helped u,

pls put thanks

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Are 2/5 4/10 6/15 and 8/20 proportional?

Over [174]

Yes, since you can multiply 2/5 numerator and denominator to get the rest of the fractions, it makes them all equivalent.

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3 years ago

Read 2 more answers

Find the equation for the plane that contains the line x=−1+3t , y=1+2t, z=2+4t and is perpendicular to the plane containing the

Ivan

Let $L$ be the line given by the vector equation

$(-1,1,2) + \lambda(3,2,4) \ , \lambda \in \mathbb{R}$ .

First, we use the director vectors of the lines L1 and L2 to get the

vector equation of the plane containing them, which we denote by $\Pi_1$ . This is,

$\\\\\Pi_1 : (1,-1,5) + \alpha (2,-1,6) + \beta (1,1,-3) \ , \alpha, \beta \in \mathbb{R}\\\\\\$

We observe that $\vec{N} = (2,-1,6)\times(1,1,-3) = (-3,12,3) \ne (0,0,0)$ . Therefore, the vector equation of $\Pi_1$ defines a plane and $\vec{N}$ is a normal vector to $\Pi_1.$

Finally, the vector equation for the wanted plane, which we denote by $\Pi$ , is

$\Pi : (-1,1,2) + r(3,2,4) + s(-3,12,3), r,s \in \mathbb{R} \ .$

Thus, if $s = 0$ , then $L \subset \Pi$ and since $\vec{N}$ is parallel to $\Pi$ , then it is perpendicular to $\Pi_1$ .

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3 years ago

Which of the following is NOT equivalent to 22/7?

jasenka [17]

Answer:

the option c is the answer for this question

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3 years ago

Area and explanation?

Olin [163]

There are a few ways to go about doing this, but the easiest is portably to break it up into three shapes: two identical triangles, and a rectangle. The area of one triangle is:
1/2 (10x8) = 40, so the area of two triangles is 80.

The area of the square is 20 x 8, which is 160.

The total area is the sum of the two:
80 + 160 = 240, and this is your total area.

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3 years ago

Given: 3x - 2 ≤ 2x + 1.

Natasha2012 [34]

Answer:

$\boxed{\boxed{\pink{\bf \leadsto Hence \ the\ correct \ option \ is \ 2 . }}}$