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3 years ago

Can y’all please answer

Mathematics

2 answers:

Andreyy893 years ago

3 0

The answer is F
Ask me anything if you still have questions

inysia [295]3 years ago

3 0

The answer is F :)

...

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Let a be an mn matrix, and let u and v be vectors in irn with the property that au = 0 and av = 0. explain why a(u + v) must be

tatiyna

$\mathbf{Au}=\mathbf0$
$\mathbf{Av}=\mathbf0$

$\implies\mathbf{Au}+\mathbf{Av}=\mathbf0$
$\implies\mathbf A(\mathbf u+\mathbf v)=\mathbf0$

Multiplying by scalars $c,d$ won't change this fact.

4 0

4 years ago

Si la variable es cuantitativa, proporcione los valores posibles o el rango de valores que toma la variable.

Genrish500 [490]

Answer:

Illegal Immigrant

Step-by-step explanation:

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4 0

4 years ago

A restaurant manager can spend at most $600 a day for operating costs and payroll. It costs $100 each day to operate the bank an

navik [9.2K]

Option A

The restaurant Manager can afford at most 10 employees for the day

Solution:

Given that restaurant manager can spend at most $600 a day for operating costs and payroll

It costs $100 each day to operate the bank and $50 dollars a day for each employee

The given inequality is:

$50x + 100\leq 600$

Where , "x" is the number of employees per day

Let us solve the inequality for "x"

$50x + 100\leq 600$

Add -100 on both sides of inequality

$50x + 100 - 100\leq 600 - 100\\\\50x\leq 500$

Divide by 50 on both sides of inequality

$\frac{50x}{50}\leq \frac{500}{50}\\\\x\leq 10$

Hence the restaurant Manager can afford at most 10 employees for the day

Thus option A is correct

6 0

4 years ago

Find the value of y (angles)

KatRina [158]

Answer:

Step-by-step explanation:

y+80 = 3y

80 = 2y

y = 40

8 0

3 years ago

Read 2 more answers

In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.

WARRIOR [948]

Answer:

For First Solution: $y_1(t)=e^t$

$y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution: $y_2(t)=cosht$

$y_2(t)=cosht$ is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: $y_1(t)=e^t$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_1(t)=e^t$

First order derivative:

$y'_1(t)=e^t$

2nd order Derivative:

$y''_1(t)=e^t$

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence $y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution:

$y_2(t)=cosht$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_2(t)=cosht$

First order derivative:

$y'_2(t)=sinht$

2nd order Derivative:

$y''_2(t)=cosht$

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence $y_2(t)=cosht$ is the solution of equation y''-y=0.

3 0

3 years ago