Which figures below are similar? 1 and 2 2 and 3 2 and 4 1 and 4

The depth of water near a boat dock is collected by a buoy. The data shows that the water level reached 11 feet during high tide

Scorpion4ik [409]

9514 1404 393

Answer:

9.17 ft

Step-by-step explanation:

Put x=12 in the equation and evaluate. (The sine argument is in degrees.)

y = 1.5·sin(25.72(12 +1.5)) +9.5 = 1.5·sin(347.22) +9.5

y = -0.33 +9.5 = 9.17

The water will be 9.17 feet deep at noon.

3 0

3 years ago

Hard question but would help me out

Alekssandra [29.7K]

Answer:

Critical value f(1)=2.

Minimum at (1,2), function is decreasing for $0$ and increasing for $x>1.$

$\left(3,\dfrac{4}{\sqrt{3}}\right)$ is point of inflection.

When 0<x<3, function is concave upwards and when x>3, , function is concave downwards.

Step-by-step explanation:

1. Find the domain of the function f(x):

$\left\{\begin{array}{l}x\ge 0\\x\neq 0\end{array}\right.\Rightarrow x>0.$

2. Find the derivative f'(x):

$f'(x)=\dfrac{(x+1)'\cdot \sqrt{x}-(x+1)\cdot (\sqrt{x})'}{(\sqrt{x})^2}=\dfrac{\sqrt{x}-\frac{x+1}{2\sqrt{x}}}{x}=\dfrac{2x-x-1}{2x\sqrt{x}}=\dfrac{x-1}{2x^{\frac{3}{2}}}.$

This derivative is equal to 0 at x=1 and is not defined at x=0. Since x=0 is not a point from the domain, the crititcal point is only x=1. The critical value is

$f(1)=\dfrac{1+1}{\sqrt{1}}=2.$

2. For $0$ the derivative f'(x)<0, then the function is decreasing. For $x>1,$ the derivative f'(x)>0, then the function is increasing. This means that point x=1 is point of minimum.

3. Find f''(x):

$f''(x)=\dfrac{(x-1)'\cdot 2x^{\frac{3}{2}}-(x-1)\cdot (2x^{\frac{3}{2}})'}{(2x^{\frac{3}{2}})^2}=$

$=\dfrac{2x^{\frac{3}{2}}-2(x-1)\cdot \frac{3}{2}x^{\frac{1}{2}}}{4x^3}=\dfrac{2x^{\frac{3}{2}}-2\cdot\frac{3}{2}x^{\frac{3}{2}}+ 2\cdot\frac{3}{2}x^{\frac{1}{2}}}{4x^3}=$

$=\dfrac{-x+3}{4x^{\frac{5}{2}}}.$

When f''(x)=0, x=3 and $f(3)=\dfrac{3+1}{\sqrt{3}}=\dfrac{4}{\sqrt{3}}.$

When 0<x<3, f''(x)>0 - function is concave upwards and when x>3, f''(x)>0 - function is concave downwards.

Point $\left(3,\dfrac{4}{\sqrt{3}}\right)$ is point of inflection.

4 0

4 years ago

I need help with this problem quick.

KatRina [158]

The absolute value - or distance from zero, is fairly simple to find. Make your final answer positive to find it.

$|\frac{10-7}{4}| = |\frac{3}{4}| = |.75| = .75$

$|\frac{-15+8}{6}| = |-\frac{7}{6}| = |-1.2| = 1.2$

8 0

4 years ago

In the triangle below, the value of a is 12.6. True False

aev [14]

Answer:

True

Step-by-step explanation:

Formula

Tan(x) = opposite / adjacent

Givens

x = 42

adjacent = 14

opposite = ?

Solution

Tan(42) = opp/14 Find tan(42)

Tan(42) = 0.900404

0.900404 = opp/ 14 Multiply both sides by 14

14 * 0.900404 = opp/14 * 14 Simplify the left

12.6 rounded = opp

a = 12.6

7 0

3 years ago

In a class in which the final course grade depends entirely on the average of four equally weighted 100-point tests, Joyce has s

d1i1m1o1n [39]

Answer:

66 ≤ f ≤100

Explanation

Mean= ( Σ x ) / n

Mean= sum of scores/ number of subject she took

Now, she already too 3 subject which sum is 85+83+86=254

Now we need to know range of score for her to have (grade) a mark between 80 and 89

Now let take the lower limit mean=80

The lowest score she can get is

Mean = ( Σx) / n

80=(85+83+86+f)/4

80×4= 254+f

Therefore, f= 320-254=66

Therefore the minimum score she can have to have a B is 66.

Then, let take the upper limit mean 89. i.e the maximum she can have so that she don't have an A grade.

Mean = ( Σx) / n

89=( 83+85+86+f)/4

89×4= 254+f

f= 356-254

f=102.

Therefore this shows that she cannot have an A grade in the exam. The maximum score for the exam is 100.

There the range of score is 66 ≤ f ≤100 to have a B grade

66 ≤ f ≤100 answer

Since she cannot score 102 in the examination.

8 0

3 years ago