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3 years ago

The temperature was 50°F before it started falling. It fell 2°F every hour for 3 hours. What

Mathematics

2 answers:

lianna [129]3 years ago

4 0

It fell 2°F every hr and 3 hrs passed meaning it went down 6°F leaving you at 44°F

Weird_Child3 years ago

0 0

44 degrees F

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Write an equation for the linear function whose graph contains the points (-1 ,-2) and (3, 10)?

Fofino [41]

Answer:

y+2 =3(x+1)

Step-by-step explanation:

Firstly I will find slope of this linear function:

m=(y-yo) / (x-xo)

m=(-2-10)/(-1-3)

m=-12/-4

m=3

Replacing A(-1-2) :

(y-yo)=m(x-xo)

3 0

3 years ago

How to solve this equation

erica [24]

| x - 2 | + 4 < 7
-7< x - 2 + 4 < 7
-7 < x + 2 < 7
-9 < x < 5

6 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

eduard

The Lagrangian is

$L(x_1,\ldots,x_n,\lambda_1,\ldots,\lambda_n)=x_1+\cdots+x_n+\lambda_1({x_1}^2+\cdots+{x_n}^2)+\cdots+\lambda_n({x_1}^2+\cdots+{x_n}^2)$

with partial derivatives (set equal to 0)

$\dfrac{\partial L}{\partial x_i}=1+2x_i(\lambda_1+\cdots+\lambda_n)=0$

$\dfrac{\partial L}{\partial\lambda_i}={x_1}^2+\cdots+{x_n}^2-36=0$

for each $1\le i\le n$ .

Let $\Lambda$ be the sum of all the multipliers $\lambda_i$ ,

$\Lambda=\displaystyle\sum_{k=1}^n\lambda_k=\lambda_1+\cdots+\lambda_n$

We notice that

$x_i\dfrac{\partial L}{\partial x_i}=x_i+2{x_i}^2\Lambda=0$

so that

$\displaystyle\sum_{i=1}^nx_i\dfrac{\partial L}{\partial x_i}=\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0$

We know that $\sum\limits_{i=1}^n{x_i}^2=36$ , so

$\displaystyle\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0\implies\sum_{i=1}^nx_i=-72\Lambda$

Solving the first $n$ equations for $x_i$ gives

$1+2\Lambda x_i=0\implies x_i=-\dfrac1{2\Lambda}$

and in particular

$\displaystyle\sum_{i=1}^nx_i=-\dfrac n{2\Lambda}$

It follows that

$-\dfrac n{2\Lambda}+72\Lambda=0\implies\Lambda^2=\dfrac n{144}\implies\Lambda=\pm\dfrac{\sqrt n}{12}$

which gives us

$x_i=-\dfrac1{2\left(\pm\frac{\sqrt n}{12}\right)}=\pm\dfrac6{\sqrt n}$

That is, we've found two critical points,

$\pm\left(\dfrac6{\sqrt n},\ldots,\dfrac6{\sqrt n}\right)$

At the critical point with positive signs, $f(x_1,\ldots,x_n)$ attains a maximum value of

$\displaystyle\sum_{i=1}^nx_i=\dfrac{6n}{\sqrt n}=6\sqrt n$

and at the other, a minimum value of

$\displaystyle\sum_{i=1}^nx_i=-\dfrac{6n}{\sqrt n}=-6\sqrt n$

4 0

4 years ago

What is the value of 3/5 x - 4 when x=5

lisov135 [29]

(3/5)(5) - 4

3 - 4

This equation is equivalent to -1 when x is equal to 5.

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

6 0

3 years ago

Click on photo! Show all work needing help ASAP

telo118 [61]

-x-y=-9

-3x-2y=-28

eliminate the x first

multiply the first system of equations by 3

(3)(-x)=-3x

(3)(-y)=-3y

(3)(-9)=-27

-3x-3y=-27

-3x-2y=-28

eliminate the systems of equations by subtracting

-3x--3x-3y--2y=-27--28

-3x+3x-3y+2y=-27+28

-y=1

multiply -1 for -y=-1 to get positive y

(-y)(-1)=(1)(-1)

y=-1

substitution method by using y=-1 into one of the system of equations

-x-y=-9

-x--1=-9

-x+1=-9

move +1 to the other side

sign changes from +1 to -1

-x+1-1=-9-1

-x=-10

multiply by -1 for -x=-10 to get positive x

(-x)(-1)=(-10)(-1)

x=10

answer:

x=10, y=-1

7 0

3 years ago