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schepotkina [342]
3 years ago
8

What is the best estimate for the product of 289 and 7

Mathematics
1 answer:
Alja [10]3 years ago
7 0
The best estimate would be 2,000 but, the original answer is 2,023
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A pet store sold 49 siamese cats. If the ratio of siamese cats to house cats sold was 7:4, what is the combined amount of cats s
S_A_V [24]

Answer:

uyygfhuittyuuuujhhtyyuiiyyuuu

4 0
3 years ago
10(8r+23) =1430 <br> HELP PLS ITS FOR MY HW PLS PLS PLS
Mademuasel [1]

Answer:

r = 15

Step-by-step explanation:

10(8r + 23) = 1430

Divide both sides by 10.

8r + 23 = 143

Subtract 23 from both sides.

8r = 120

Divide both sides by 8.

r = 15

7 0
3 years ago
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
A volleyball reaches its maximum of 13 feet, 3 seconds after it is served. What quadratic formula could model the height of the
Katena32 [7]

Answer:

H=-(t-3)^2+13

Step-by-step explanation:

The path covered by the volleyball will be a downward parabola with the vertex being the highest point of the ball.

A general form of a downward parabola is given as:

y=-a(x-h)^2+k

Where (h,k) is the vertex of the parabola and 'a' is a constant.

Now, let 'h' be the vertical height and 't' be the time taken.

So, the equation would be of the form:

H=-a(t-h)^2+k

Now, as per question:

h = 2 seconds, k = 13 feet.

H=-a(t-3)^2+13

Now, taking a = 1. So, the formula that can be used is:

H=-(t-3)^2+13

7 0
3 years ago
Create a linear function that has a solution at (-2,6)
xenn [34]
(x,y) so you start at the orgin and go down two because it's negative. then you go right six and plot your point.

negative number - down or left

positive number - up or right
8 0
3 years ago
Read 2 more answers
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