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2 years ago

Find the area AC = 8m, BD = 6m

Mathematics

1 answer:

grandymaker [24]2 years ago

5 0

Answer:

24 m²

Step-by-step explanation:

The area of a rhombus is the product of the two diagonals over 2

A= (8*6)/2 = 24 m²

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For some reason, in our math class

Alona [7]

Then there would be 9 girls and 23 boys

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3 years ago

PLEASE HELP NOBODY KNOWS ? Mhmh

Sholpan [36]

Answer:

76%

Step-by-step explanation:

Times the decimal by 100 to get a percent

basically moving the decimal 2 spaces to the right.

4 0

2 years ago

If i earn 165 dollars in 20 hours how many would i make in 40 hourr

Triss [41]

Answer:

330

Step-by-step explanation:

165 times 2 equals 312. This is because you're doubling the amount of money and hours.

7 0

2 years ago

If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn

mash [69]

Let $a=\tan A$ and $b=\sin A$ . Then

$m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab$

$\implies m^2-n^2=4\tan A\sin A$

and

$mn=(a+b)(a-b)=a^2-b^2$

$\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}$

The expression under the square root can be rewritten as

$\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)$

Recall that

$\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A$

so that

$\tan^2A-\sin^2A=\sin^2A\tan^2A$

and assuming $\sin A>0$ and $\tan A>0$ , we end up with

$4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A$

so that

$m^2-n^2=4\sqrt{mn}$

as required.

5 0

2 years ago

125+25x = 50 I need a answer quickly

antoniya [11.8K]

Answer:

-3

Step-by-step explanation:

Subtract 25x 125=-25x+50

Subtract 50 125-50= -25x

Simplify 75= -25x

Divided 75 by negatrive 25

8 0

2 years ago