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aniked [119]
2 years ago
12

PLEASE HELP ME YOU GET IT RIGHT I WILL MARK BRAINLIST but if its wrong... i report you

Mathematics
1 answer:
mamaluj [8]2 years ago
6 0
I’m pretty sure it’s 60 but I’m not 100% sure so I’m sorry if it’s wrong
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Solve the system of equations. If the system has no solution, say that it is inconsistent. Graph the lines of the system.
Romashka-Z-Leto [24]

Answer:

The solution to the system of the equations will be:

x=3,\:y=1

Step-by-step explanation:

Given the system of the equations

x+y=4

x-y=2

solving the system of the equations

\begin{bmatrix}x+y=4\\ x-y=2\end{bmatrix}

x-y=2

-

\underline{x+y=4}

-2y=-2

solving

-2y=-2

Divide both sides by -2

\frac{-2y}{-2}=\frac{-2}{-2}

y=1

\mathrm{For\:}x+y=4\mathrm{\:plug\:in\:}y=1

x+1=4

x=3

Therefore, the solution to the system of the equations will be:

x=3,\:y=1

5 0
3 years ago
What number completes the following equation 8× (40+7) =(8× )+(×7)
Vladimir [108]
8x( 47 ) = 8x + 7x
8x( 47 ) = 8x( 2 ) - x
8x( 47 - 2 ) = - x
8x( 45 ) = - x
360x = - x
x = -360x

I don't think the question can be done
Is the question wrong?
because there's no answer other than 0.
5 0
3 years ago
Please help me due at tomorrow 8a.m please
laiz [17]

Answer:

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What is the upper bound of the function f(x)=4x4−2x3+x−5?
inessss [21]

Answer:

(no global maxima found)

Step-by-step explanation:

Find and classify the global extrema of the following function:

f(x) = 4 x^4 - 2 x^3 + x - 5

Hint: | Global extrema of f(x) can occur only at the critical points or the endpoints of the domain.

Find the critical points of f(x):

Compute the critical points of 4 x^4 - 2 x^3 + x - 5

Hint: | To find critical points, find where f'(x) is zero or where f'(x) does not exist. First, find the derivative of 4 x^4 - 2 x^3 + x - 5.

To find all critical points, first compute f'(x):

d/( dx)(4 x^4 - 2 x^3 + x - 5) = 16 x^3 - 6 x^2 + 1:

f'(x) = 16 x^3 - 6 x^2 + 1

Hint: | Find where f'(x) is zero by solving 16 x^3 - 6 x^2 + 1 = 0.

Solving 16 x^3 - 6 x^2 + 1 = 0 yields x≈-0.303504:

x = -0.303504

Hint: | Find where f'(x) = 16 x^3 - 6 x^2 + 1 does not exist.

f'(x) exists everywhere:

16 x^3 - 6 x^2 + 1 exists everywhere

Hint: | Collect results.

The only critical point of 4 x^4 - 2 x^3 + x - 5 is at x = -0.303504:

x = -0.303504

Hint: | Determine the endpoints of the domain of f(x).

The domain of 4 x^4 - 2 x^3 + x - 5 is R:

The endpoints of R are x = -∞ and ∞

Hint: | Evaluate f(x) at the critical points and at the endpoints of the domain, taking limits if necessary.

Evaluate 4 x^4 - 2 x^3 + x - 5 at x = -∞, -0.303504 and ∞:

The open endpoints of the domain are marked in gray

x | f(x)

-∞ | ∞

-0.303504 | -5.21365

∞ | ∞

Hint: | Determine the largest and smallest values that f achieves at these points.

The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:

The open endpoints of the domain are marked in gray

x | f(x) | extrema type

-∞ | ∞ | global max

-0.303504 | -5.21365 | global min

∞ | ∞ | global max

Hint: | Finally, remove the endpoints of the domain where f(x) is not defined.

Remove the points x = -∞ and ∞ from the table

These cannot be global extrema, as the value of f(x) here is never achieved:

x | f(x) | extrema type

-0.303504 | -5.21365 | global min

Hint: | Summarize the results.

f(x) = 4 x^4 - 2 x^3 + x - 5 has one global minimum:

Answer: f(x) has a global minimum at x = -0.303504

5 0
3 years ago
Read 2 more answers
Please Help me!!! <br><br><br> I know the answer but I don’t understand how to get that answer
worty [1.4K]
-7 ÷ 2 1/2
Okay, so first, put the -7 over one and make 2 1/2 an improper fraction. so now:
-7/1 ÷ 5/2
Then, take the reciprocal of 5/2 and multiply it by -7/1
-7/1 × 5/2
Which would be -35/2
5 0
3 years ago
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