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3 years ago

Find the midpoint of the line segment with the given endpoints. (-1,0), (-3,-4)

Mathematics

2 answers:

larisa86 [58]3 years ago

5 0

No can’t u just search it up I feel like you’ll get the answer to this question in to time

Gre4nikov [31]3 years ago

5 0

Answer:

(-2,-2)

Step by step:

(-1,0) (-3,-4)

(xa,ya)(xa,yb)

$\frac{ x ^{a} + {x}^{b} }{2} . \frac{y ^{a} + y^{b} }{2} \\ = \frac{ - 1 + - 3}{2} . \frac{0 + - 4}{2} \\ = \frac{ - 4}{2} \frac{ - 4}{2}$

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Could someone help me pleaseee

Savatey [412]

Answer:

a) We are given the length of both legs. We need to find the hypotenuse.

b) 40^2 + 71.5^2 = X^2

c) 40^2 + 71.5^2 =

1600 + 5,112.25 = 6,712.25 in.

d) 6,712 inches

Step-by-step explanation:

8 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

SIR MELIODAS CAPTION MEL|ODAS

dalvyx [7]

Answer:

great

Step-by-step explanation:

5 0

3 years ago

This graph shows a proportional relationship. What is the constant of proportionality? Enter your answer as a ratio in simplifie

AnnyKZ [126]

The answer is 5/4. I just took the test.

4 0

3 years ago

Read 2 more answers

PLEASE HELP

Tems11 [23]

QUESTION A

The given multiplication problem is

$\frac{39}{64} \times \frac{8}{13}$

Factor each term to obtain;

$\frac{13\times 3}{8\times8} \times \frac{8}{13}$

Cancel out the common factors to obtain;

$\frac{1\times 3}{8\times1} \times \frac{1}{1}$

Simplify to get;

$\frac{3}{8}$

QUESTION B

The given multiplication problem is

$\frac{2}{3}\times \frac{1}{5}\times \frac{4}{7}$

This the same as

$\frac{2\times 1\times 4}{3\times 5\times 7}$

This simplifies to;

$\frac{8}{105}$

QUESTION C

The given problem is

$\frac{3}{5}\times \frac{10}{12} \times \frac{1}{2}$

This is the same as

$\frac{3}{5}\times \frac{5}{6} \times \frac{1}{2}$

$=\frac{1}{1}\times \frac{1}{2} \times \frac{1}{2}$

This simplifies to

$=\frac{1}{4}$

QUESTION D.

The given expression is

$\frac{4}{9}\times 54$

Factor the 54 to obtain;

$\frac{4}{9}\times 9\times 6$

Cancel the common factors to get;

$\frac{4}{1}\times 1\times 6$

This simplifies to;

$=24$

QUESTION E

The given problem is

$20\times 3\frac{1}{5}$

Convert the mixed numbers to improper fraction to obtain;

$=20\times \frac{16}{5}$

$=4\times5 \times \frac{16}{5}$

Cancel the common factors to get;

$=4\times1 \times \frac{16}{1}$

$=64$

QUESTION F

The multiplication problem is

$11 \times 2 \frac{7}{11}$

Convert the mixed numbers to improper fractions to obtain;

$11 \times \frac{29}{11}$

Cancel out the common factors to get;

$=1 \times \frac{29}{1}$

Simplify;

$=29$

QUESTION G

The given problem is

$5\frac{1}{3}\times 5\frac{1}{8}$

Convert to improper fractions;

$=\frac{16}{3}\times \frac{41}{8}$

Cancel out the common factors to get;

$=\frac{2}{3}\times \frac{41}{1}$

$=\frac{82}{3}$

Convert back to mixed numbers

$=27\frac{1}{3}$

QUESTION H

The given expression is

$10\frac{2}{3} \times 1\frac{3}{8}$

Convert to improper fraction to get;

$\frac{32}{3} \times \frac{11}{8}$

Cancel common factors to get;

$=\frac{4}{3} \times \frac{11}{1}$

Simplify

$=\frac{44}{3}$

Convert back to mixed numbers;

$=14\frac{2}{3}$

7 0

3 years ago