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3 years ago

The area of the irregular polygon is ____ square units.

Mathematics

1 answer:

vova2212 [387]3 years ago

4 0

15 square units. Its right

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30= 2w + 2h<br><br> W= 3/4 h

Alexeev081 [22]

Answer:

h=60/7

w=45/7

Step-by-step explanation:

just try it.

7 0

3 years ago

Read 2 more answers

Write each expression without the absolute value symbol |3x+12|

klasskru [66]

Answer:

Step-by-step explanation:

|3x+12|

one is as it is one is the negative value of it

3x + 12 - (3x + 12)

- 3x - 12

That is because both negative and positive values put in am absolute value become positive.

3 0

2 years ago

If you need help- an anything just and MATH OR SIENT

galben [10]

It won’t let me add:(

3 0

2 years ago

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$

3 0

2 years ago

A store can buy 3 pairs of shorts for $10 . How much would the store pay for a dozen

FromTheMoon [43]

Answer:

$40

Step-by-step explanation:

12 ÷ 3 = 4

4 × 10 = 40

So, the answer is $40.

<h2><u><em>Please mark as Brainliest!!!</em></u></h2>

4 0

2 years ago