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3 years ago

9

How to determine if bacteriostatic or bactericidal?

1 answer:

kaheart [24]3 years ago

4 0

The way to determine bacteriostatic or bacterial is by using a micorscope

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A grocery store gives away a $10 gift card to every 25th customer and a $20 gift card to every 60th customer. a. Which customer

steposvetlana [31]

To find the answer you need to find the common factor of numbers 25 and 60

25,50,75,100,125,150,175,200,225,250,275,300

60,120,180,240,300

The 300th customer will receive both cards, and a total of $220 and 17 gift cards were given away.

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4 years ago

1.35 x 8.8 please send a picture step by step I’m very confused thank you

Igoryamba

Answer:

Check out the image below!

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

After one hour, the hare had finished 2/3 of a 100-yard race. In that same time, the tortoise had finished 42 3/4 yards. How muc

andriy [413]

Answer:

23 11/12 yards or 23.9166 yards

Step-by-step explanation:

2/3 of 100 yard race is equivalent to 66 2/3 yards. Tge difference between the above and 42 3/4 yards covered by tortoise will be

66 2/3- 42 3/4= 23 11/12 yards or 23.91666 yards

5 0

3 years ago

Find the y-intercept and the slope of the line<br><br> y=-x+7<br><br> y-intercept:<br> slope:

Ipatiy [6.2K]

Answer:

The y-intercept would be 7

The slope would be -1 or -1/1. They are both the same thing.

The slope is the number which is located directly to the left of the variable (the variable in this case is x, but in other situations it could be a totally different letter).

5 0

3 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago