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2 years ago

In Bear Creek Bay in July, high tide is at 1:00 pm. The water level at

1 answer:

5 0

The reason we use cosine, is because a cosine graph is at its maximum at the beginning of the cycle.
We begin with basic cosine equation:
Y(x)=Acos(Bx+c)+D
First we need to determine the middle line of the graph (amplitude) by finding the difference between tides and dividing by two: (7-1)/2=3
We know that period is 12. This means that b needs to be pi/6 or <span>0.523 </span>if 12b=2pi
Then we need to average out the tides, to determine how much the curve needs to shift: (7+1)/2= 4
In addition, we need to shift the equation to the right since high tide is at 1pm or 13 hours after midnight.
In the end we get the following equation:
y=4+3cos(0.523(x-13))

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The function f(x) = x2 has been translated 9 units up and 4 units to the right to form the function g(x). Which represents g(x)?

Alenkinab [10]

For this case, the parent function is given by:

$f (x) = x ^ 2$

We apply the following transformations:

Vertical translations:

Suppose that k> 0

To graph y = f (x) + k, move the graph of k units upwards:

For k = 9 we have:

$h (x) = x ^ 2 + 9$

Horizontal translations:

Suppose that h> 0

To graph y = f (x-h), move the graph of h units to the right

For h = 4 we have:

$g (x) = (x-4) ^ 2 + 9$

Answer:

The function g (x) is given by:

$g (x) = (x-4) ^ 2 + 9$

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2 years ago

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A lotion is made from an oil blend costing $1.50 per ounce and glycerin costing $1.00 per ounce. Four ounces of lotion costs $5.

Svet_ta [14]

Answer:

A (4-g)1.5

Step-by-step explanation:

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2 years ago

A student spends 17/35 of his pocket money on transport and fruits. He spends 5/6 of the remainder on sweets. What fraction of h

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Answer: 3/7 is the pocket money he spends on sweets

Step-by-step explanation:

Let the amount student spends be x

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2 years ago

WILL GIVE BRAINLIEST ANSWER!!

Natasha2012 [34]

Check the picture below.

now, bear in mind that, there are 60minutes in 1 degree, so 23' is just 23/60 degrees or about 0.38 degrees.

and 29' is just 29/60 degrees or about 0.48 degrees.

so 16°23' is about 16.38°, and 49°29' is about 49.48°

$\bf tan(16.38^o)=\cfrac{200}{a+b}\implies a+b=\cfrac{200}{tan(16.38^o)}
\\\\\\
\boxed{b=\cfrac{200}{tan(16.38^o)}-a}
\\\\\\
tan(49.48^o)=\cfrac{200}{b}\implies \boxed{b=\cfrac{200}{tan(49.48^o)}}\\\\
-------------------------------\\\\
\cfrac{200}{tan(49.48^o)}=\cfrac{200}{tan(16.38^o)}-a\\\\\\a=\cfrac{200}{tan(16.38^o)}-\cfrac{200}{tan(49.48^o)}$

make sure your calculator is in Degree mode.

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