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Ivanshal [37]
3 years ago
11

The sum of two numbers is -25. One number is 101 less than the other. Find the numbers

Mathematics
2 answers:
zlopas [31]3 years ago
6 0

Answer:

-63, 38

Step-by-step explanation:

1st number = x

2nd number = x+101

x+x+101=-25

2x=-126

x=-63

1st number = x = -63

2nd number = x+101 = 38

LuckyWell [14K]3 years ago
6 0

Answer:

x=38

y= -63

Step-by-step explanation:

x+y=-25

y=x-101

x+(x-101)= -25

2x=-25+101

2x=76

x=76/2

x=38

y=x-101

y=38-101

y= -63

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PLZ HELP NEED TWO ANSWERS
densk [106]

Answer:

50 from home to school and 55 from school to home.

Step-by-step explanation:

55 is 5 more then 50.

6 0
3 years ago
For every batch of cookies Carla's made , her brother ate 2 cookies. If Carla's brother ate a total of 6 cookies, how many batch
spayn [35]

Step-by-step explanation:

number of the total cookies ÷ the nber her brother ate

6÷2=3

3 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
What is the following product
dangina [55]

Option B is correct.

Step-by-step explanation:

We need to solve: \sqrt[3]{x^2}\sqrt[4]{x^3}

We know that: \sqrt[n]{x}\sqrt[b]{x} =\sqrt[n*b]{x.x}= \sqrt[n*b]{x^2}

Applying the above rule:

\sqrt[3]{x^2}\sqrt[4]{x^3}\\=\sqrt[3*4]{x^2.x^3}\\=\sqrt[12]{x^5}

So, Option B is correct.

Keywords: Solving with Exponents

Learn more about Solving with Exponents at:

  • brainly.com/question/4934417
  • brainly.com/question/13174254
  • brainly.com/question/13174255

#learnwithBrainly

3 0
3 years ago
Noon to 3:05 elapsed time
GrogVix [38]
Noon is 12, I'm guessing you mean both of these at am, or both at pm. 12-3 is 3 hours, but there's 5 extra, so the elapsed time would be 3:05.
6 0
3 years ago
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