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DENIUS [597]
3 years ago
6

16 + 2q = 6 what’s this?

Mathematics
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

6 - 16 = -10

-10/2 = -5

q = -5

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PLEASE HELP ASAP!!!!!
erastova [34]

Answer:

B = (3, 3)

Step-by-step explanation:

Since the line segment is fifteen units long, from the looks of it, chopping ⅓ of A's side, led to the ordered pair of (3, 3).

I hope this helps you out alot, and as always, I am joyous to assist anyone at any time.

7 0
3 years ago
Simplify the following expression into the form a + bi, where a and b are rational numbers.
vagabundo [1.1K]
The answer is -16 - 10i.

Using the distributive property on the first part, we have:
-2i*7--2i*4i + (3+i)(-2+2i)
-14i+8i² +(3+i)(-2+2i)

Using FOIL on the last part,
-14i+8i²+(3*-2+3*2i+i*-2+i*2i)
-14i+8i²-6+6i-2i+2i²
-10i+8i²-6+2i²

Since we know that i = -1,
-10i+8(-1)-6+2(-1)
-10i-8-6-2
-16-10i
7 0
3 years ago
Read 2 more answers
An astronaut is 60kg on the Moon. What is his Mass on Earth?
Rus_ich [418]

Answer:

10kg?

Step-by-step explanation:

3 0
3 years ago
Four balls of wool will make 8 knitted caps. How many balls of wool will malcolm need if he wants to make six caps?
Olenka [21]
I could be wrong but I think it is 2
7 0
3 years ago
Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
7 0
3 years ago
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