All categories

3 years ago

–42t + 80t + –59t − –97 = –29

Mathematics

2 answers:

Rom4ik [11]3 years ago

8 0

Answer:

The answer is t= -2

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

Hoped this helped!

Could I perhaps get brainly, please?

Aleksandr [31]3 years ago

5 0

Answer:

Hope this helps

Step-by-step explanation:

t=6

You might be interested in

Which statement is true?

Mumz [18]

The answer is D.

|-4| < 4
4 < 4

|-1| < 0
1 < 0

|-6| = -6
6 = -6

|-14| > 10
14 > 10

6 0

3 years ago

A square has one side that measures 6 inches. What is the perimeter of the square?

MariettaO [177]

Answer:

24 inches

Step-by-step explanation:

a square has 4 equal sides. if one side is 6, the rest of the sides are also six. perimeter is all the sides added. so 6+6+6+6, or 6*4, equals 24 inches as perimeter.

4 0

3 years ago

Did it rain spaghetti more or less than ½ of the month? Explain how you know. pls explain

olga55 [171]

Answer:

More and Brainliest PLS

Step-by-step explanation:

there is 20 spaghetti days

half of 30 is 15

20 is greater than 15 so it is more than half of the month

8 0

4 years ago

You deposit $7200 in an account that pays 9.3% interest compounded yearly. Find the balance after 9 years.

inysia [295]

Answer:

Step-by-step explanation:

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