Find two numbers whose difference is 10 and whose product is a minimum

How do I find the simplest form of a square root?

lozanna [386]

Answer:

A General Note: The Product Rule for Simplifying Square Roots. If a and b are nonnegative, the square root of the product a b \displaystyle ab ab is equal to the product of the square roots of a and b.

Step-by-step explanation:

3 0

3 years ago

1- Emily collected $950 selling Girl Scout cookies all day Saturday. Emily's Troop collected 10 times as much as she did. How mu

VikaD [51]

1. $9500 because 10*950=9500
2. $95 because 950/10=95

( can explain further )

5 0

3 years ago

Thanks for your answer

pogonyaev

I just had a test on this lol perfect timing. Anyways the answer would be 3 times bigger bc the circumference of a circle with the radius of 3 is 6pi and the circumference of a circle with the radius of 1 is 2pi which makes the first circle 3 times bigger

5 0

3 years ago

What are equivalent fraction to 33/48 with the numeration as 11

Vera_Pavlovna [14]

Answer:

11/16

Step-by-step explanation:

The first thing I would do is see 11 times what will get me 33

The answer is 3

Next I would ask myself 48 divided by 3 is what

The answer is 16

To double check my work I would do 3 x 11 = 33

3 x 16 = 48

5 0

3 years ago

If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =

attashe74 [19]

First of all, let θθ be some angle in (0,π)(0,π). Then

θθ is acute ⟺⟺ θ<π2θ<π2 ⟺⟺ cosθ>0cos⁡θ>0.θθ is right ⟺⟺ θ=π2θ=π2 ⟺⟺ cosθ=0cos⁡θ=0.θθ is obtuse ⟺⟺ θ>π2θ>π2 ⟺⟺ cosθ<0cos⁡θ<0.

Now, to see if (say) angle AA of the triangle ABCABC is acute/right/obtuse, we need to check whether cos∠BACcos⁡∠BAC is positive/zero/negative. But what is cos∠BACcos⁡∠BAC? It is the angle made by the vectors AB−→−AB→ and AC−→−AC→. (When you are computing the angle at a particular vertex vv, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex vv as the initial point.) We will first compute these two vectors:

AB−→−=(0,0,0)−(1,2,0)=(−1,−2,0)AB→=(0,0,0)−(1,2,0)=(−1,−2,0)AC−→−=(−2,1,0)−(1,2,0)=(−3,−1,0)AC→=(−2,1,0)−(1,2,0)=(−3,−1,0)Therefore, the angle between these vectors is given by:cos∠BAC=AB−→−⋅AC−→−|AB−→−||AC−→−|=…(1)(1)cos⁡∠BAC=AB→⋅AC→|AB→||AC→|=…Can you take it from here? From the sign of this value, you should be able to decide if angle AA is acute/right/obtuse.

Now, do the same procedure for the remaining two angles BB and CC as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

6 0

3 years ago