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3 years ago

How to solve this question

Mathematics

2 answers:

kaheart [24]3 years ago

7 0

Answer:

I think you add all the sides but I could be wrong

sasho [114]3 years ago

3 0

To find the area, cut the figure into one triangle and one rectangle.
Area=bh for the rectangle so multiply 110 times 140
Area=bh divided by 2 so multiply 90 times 70 divided by 2

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Solve for f pleaseeeee

KonstantinChe [14]

Answer:

$\frac{3}{5}f = 4-2m\\ f= \frac{(4-2m) X 5}{3} \\f= \frac{20-10m}{3}$

P.S. If u want a numerical value for 'f' , now put the value of 'm'

7 0

3 years ago

Please help! <br> Trigonometry review

charle [14.2K]

Answer: x = 9

Step-by-step explanation:

15 / 20 = (2x + 3) / 28

cross multiply

15(28) = 20(2x + 3)

420 = 40x + 60

360 = 40x

x = 9

6 0

2 years ago

Professor Vinculum investigated the migration season of the Bulbul bird from their natural wetlands to a warmer climate. He foun

coldgirl [10]

The population of the Bulbul birds at the start of the migration season is 1800.

<h3>Exponential function</h3>

An exponential function is given by:

y = abˣ

Let y, x are variables, a is the initial value and b is the multiplication factor.

Let P be the population of the bird t days since the start of the migration season.

$P=1350+400(1.25)^{-t}\\\\At\ the\ start\ of\ the\ season(t=0):\\\\P=1350+400(1.25)^{-0}\\\\P=1350+450=1800$

The population of the Bulbul birds at the start of the migration season is 1800.

Find out more on Exponential function at: brainly.com/question/12940982

8 0

2 years ago

PLEASE HELP ME WITH THIS ASAP

Arte-miy333 [17]

Eddie: $10-3=$7
Marsha: $8+$3=$11

7 0

1 year ago

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

nydimaria [60]

Answer:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

Step-by-step explanation:

Given

$\cos(\theta) = -\frac{2}{3}$

$\theta \to$ Quadrant III

Required

Determine $\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

We have:

$\cos(\theta) = -\frac{2}{3}$

We know that:

$\sin^2(\theta) + \cos^2(\theta) = 1$

This gives:

$\sin^2(\theta) + (-\frac{2}{3})^2 = 1$

$\sin^2(\theta) + (\frac{4}{9}) = 1$

Collect like terms

$\sin^2(\theta) = 1 - \frac{4}{9}$

Take LCM and solve

$\sin^2(\theta) = \frac{9 -4}{9}$

$\sin^2(\theta) = \frac{5}{9}$

Take the square roots of both sides

$\sin(\theta) = \±\frac{\sqrt 5}{3}$

Sin is negative in quadrant III. So:

$\sin(\theta) = -\frac{\sqrt 5}{3}$

Calculate $\csc(\theta)$

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

We have: $\sin(\theta) = -\frac{\sqrt 5}{3}$

So:

$\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}$

$\csc(\theta) = \frac{-3}{\sqrt 5}$

Rationalize

$\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}$

$\csc(\theta) = \frac{-3\sqrt 5}{5}$

So, we have:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

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

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)$

Substitute: $\csc(\theta) = \frac{-3\sqrt 5}{5}$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}$

Take LCM

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

6 0

2 years ago