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3 years ago

On Monday, 1 7/8 inches of rain fell. On Tuesday, it rained 1/4 Inch. What was the total rainfall for the

Mathematics

1 answer:

rodikova [14]3 years ago

6 0

Answer:

Step-by-step explanation:

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Triangle XYZ has vertexes at X(-4, 5), Y(-3, 5), and Z(-3, 4). Find the measure of angle Y to the nearest degree.

hodyreva [135]

Mark these points on the graph to form a right angled triangle.
Angle Y will be 90 degrees

8 0

3 years ago

(-w^3 + 8w^2 – 3w) + (-8w^2 + w + 3) =

kherson [118]

The expression simplified is -w^3-2w+3

I hope this helped

4 0

3 years ago

A population of rabbits in a lab, p(x), can be modeled by the function p(x) = 20(1.014)x , where x represents the number of days

Nitella [24]

Answer:

20 = initial population of the rabbits

1.014 = growth rate of the rabbits

the average rate of change from day 50 to day 100 is 0.8

Step-by-step explanation:

A population of rabbits in a lab, p(x), can be modeled by the function

p(x) = 20(1.014)^x

This model is exponential. Where 20 = initial population of the rabbits

1.014 = growth rate of the rabbits with 1.4% increase rate of the rabbits

To find the average rate of change from day 50 to day 100,

find the population p(50) and p(100). Subtract them and divide by 100 - 50 = 50.

p(50) = 20(1.014)50 = 40.08...

p(100) = 20(1.014)100 = 80.32...

(80.32 - 40.08) / (100 - 50) = 40.24/50 = 0.8048. which is approximately 0.8 to the nearest tenth.

The rate of change is 0.8.

7 0

3 years ago

AHHHH please help answer and I’ll give you points

german

Answer: Yes She is correct

Step-by-step explanation:

4 0

3 years ago

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities

$\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

The marked are the trigonometric identities

3 0

2 years ago