Ask question

Ask question

All categories

2 years ago

5

Help me please guys to solve this :(

1 answer:

Fynjy0 [20]2 years ago

4 0

Answer:

WHAT THOSE HAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

You might be interested in

Help please screenshot

Bingel [31]

48 cm. You have 16 units and know that 2 of them equal 6 cm. So you can either divide 16 by 2 and multiply by 6 or notice that each square is worth 3 cm and multiply 16 by 3 to get 48.

7 0

2 years ago

Find the missing side. Round to the nearest tenth.

Firdavs [7]

Answer: <em>23.7</em>

Step-by-step explanation:

<em>Let's calculate the acute angle at the base of the triangle:</em>

<em>180° - 90° - 40° = 50°</em>

<em>Find side x:</em>

$\dfrac{sin \: 90^{o} }{31} =\dfrac{sin \: 50^{o} }{x}\\\\sin \: 90^{o} \cdot x=sin \: 50^{o} \cdot 31\\\\sin \: 90^{o} =1\\sin \: 50^{o} \approx 0.766\\1 \cdot x =0.766 \cdot 31\\x=23.746\\23.746 \approx 23.7$

5 0

1 year ago

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 3 (*sqaure root sign*) t (20 − t), [0,

mafiozo [28]

9514 1404 393

Answer:

maximum: 15∛5 ≈ 25.6496392002
minimum: 0

Step-by-step explanation:

The minimum will be found at the ends of the interval, where f(t) = 0.

The maximum is found in the middle of the interval, where f'(t) = 0.

$f(t)=\sqrt[3]{t}(20-t)\\\\f'(t)=\dfrac{20-t}{3\sqrt[3]{t^2}}-\sqrt[3]{t}=\sqrt[3]{t}\left(\dfrac{4(5-t)}{3t}\right)$

This derivative is zero when the numerator is zero, at t=5. The function is a maximum at that point. The value there is ...

f(5) = (∛5)(20-5) = 15∛5

The absolute maximum on the interval is 15∛5 at t=5.

5 0

2 years ago

Let e f (x) = x2 – 2x - 4 What is the average rate of change for the quadratic function from X = -1 tox to x = 42 Enter your ans

sammy [17]

Answer:

The average rate of change of the given function

A(x) = 1

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given function f(x) = x² - 2x -4

And given that x = a = -1 and x=b = 4

The average rate of change of the given function

$A(x) = \frac{f(b)-f(a)}{b-a}$

<u><em>Step(ii):-</em></u>

f(x) = x² - 2x -4

f(-1) = (-1)² - 2(-1) -4 = 1+2-4 = -1

f(4) = 4² -2(4) -4 = 16 -12 = 4

The average rate of change of the given function

$A(x) = \frac{f(b)-f(a)}{b-a}$

$A(x) = \frac{4-(-1)}{4-(-1)} = \frac{5}{5} = 1$

<u><em>final answer:-</em></u>

The average rate of change of the given function

A(x) = 1

5 0

2 years ago

Read 2 more answers

What is the least common multiple of 8 and 12?<br> 24<br> 48<br> 96

Anna007 [38]

Answer:

The least common multiple of 8 and 12 is 24

Explanation:

Find and list the multiples each number until the first common multiple is found.

This is the lowest common multiple .

Multiples of 8

8, 16, <u>24</u>, 32,40

Multiples of 12

12,<u>2</u><u>4</u>, 36,48

Therefore,

LCM of 8 and 12 is 24

3 0

2 years ago

Read 2 more answers