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2 years ago

Express in simplest form -45%

Mathematics

1 answer:

Ksenya-84 [330]2 years ago

8 0

Answer: hello, your answer is -0.45

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postnew [5]

The question is asking to choose among the following choices that states the early withdrawal fee of the account given in the problem, base on my calculation and further formulation about the said problem, I would say that the answer would be letter D. $3.75. I hope you are satisfied with my answer

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2 years ago

Read 2 more answers

25 POINTS And Brainliest!!

tatuchka [14]

The appropriate measure of central tendency is one that shows

difference and is suitable for a scale that is nominal.

Response:

The measure of central tendency to use is the <u>mode</u>.

<h3>How can the appropriate measure of central tendency be selected?</h3>

The mean is the sum of the measurements divided by the number count

of the plants.

The mode is the measurement that has the highest frequency.

The median is the measurement of the middle plant when arranged in a

given order according to size.

To argue that there is a difference between the plants, the measure of

central tendency to use is the mode, given that the data involves

measurements which can be expressed in a nominal scale.

Therefore;

The measure of central tendency that will be best for Mrs. Hull to use is the<u> mode</u>

Learn more about the measures of central tendencies here:

brainly.com/question/1027437

6 0

1 year ago

A linear function contains the following points.

DIA [1.3K]

The answer is A
in a y-int your x coordinate is always going to be 0

6 0

2 years ago

Need help with this

Vesna [10]

Answer:

the answer is c

Step-by-step explanation:

the answer is c

6 0

3 years ago

Read 2 more answers

Prove :

Sauron [17]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{\sec\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

We can convert sec(α) to 1 / cos(α):

$\displaystyle \frac{1}{1/\cos\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Multiply both layers of the first fraction by cos(α):

$\displaystyle \frac{\cos\alpha}{1+\cos\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Create a common denominator. We can multiply the first fraction by (1 - cos(α)):

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Simplify:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{1-\cos^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

From the Pythagorean Identity, we know that cos²(α) + sin²(α) = 1 or equivalently, 1 - cos²(α) = sin²(α). Substitute:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{\sin^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Subtract:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Distribute:

$\displaystyle \frac{\cos\alpha-\cos^2\alpha-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Rewrite:

$\displaystyle \frac{(\cos\alpha)-(\cos^2\alpha+\cos\alpha)}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Split:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos^2\alpha+\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Factor the second fraction, and substitute sin²(α) for 1 - cos²(α):

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{1-\cos^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Factor:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{(1-\cos\alpha)(1+\cos\alpha)}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Cancel:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha}{(1-\cos\alpha)}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Divide the second fraction by cos(α):

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}=\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Hence proven.

7 0

2 years ago

Express in simplest form -45%​

Express in simplest form -45%