All categories

3 years ago

Solve the formula for the indicated variable A=1/2bh solve for h

Mathematics

1 answer:

vfiekz [6]3 years ago

7 0

Answer:

h=A*2b

Step-by-step explanation:

A=1/2bh

1/2bh=A

1/2bh÷1/2bh=A÷1/2bh (the 1/2b will cancel with the 1/2b on the left leaving h)

h=A*2b

You might be interested in

Leann is 3 cm taller than Fred. If Fred's height is f cm, then Leann's height is cm?

jek_recluse [69]

F + 3 = leann's height

6 0

3 years ago

Which expression is equivalent to 16x4 – 64? and explain the answer plzz

GuDViN [60]

Answer:

4(4x1-16)

Hope this helps

Have a good day :)

PS I disibuted 4 to 4x1 to get 16 and 4 to -16 to get -64

Step-by-step explanation:

3 0

3 years ago

Eli works in an electronics repair shop. he earns 9 dollars an hour plus a bonus for each item her repairs.

Jobisdone [24]

The equation is
a=9h+bn

3 0

3 years ago

Read 2 more answers

The figure is made from squares of different sizes.Find the area of A if the side of each of the two indentical smallest square

Temka [501]

Answer:

The area of square A is 256 cm²

Step-by-step explanation:

The four sides of a square are equal in length
Area of a square = side × side

In the given figure

∵ The sides of the square E = 2 cm each

∵ The sides of the square F = 2 cm each

∵ The side of square D is the sum of the sides of squares E and F

∴ The sides of the square D = 2 + 2 = 4 cm each

∵ The side of square C is the sum of the sides of squares E and D

∴ The sides of square C = 2 + 4 = 6 cm each

∵ The side of square B is the sum of the sides of squares C, E, and F

∴ The sides of square B = 6 + 2 + 2= 10 cm each

∵ The side of square A is the sum of the sides of squares B, F, and D

∴ The sides of square A = 10 + 2 + 4= 16 cm each

∵ Area of square A = side × side

∴ Area of square A = 16 × 16

∴ Area of square A = 256 cm²

∴ The area of square A is 256 cm²

3 0

2 years ago

Find all possible values of α+

const2013 [10]

Answer:

$\rm\displaystyle 0,\pm\pi$

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

===========================

we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )$

factor out tanγ:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)$

divide both sides by tanαtanβ-1 and that yields:

$\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}$

multiply both numerator and denominator by-1 which yields:

$\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)$

recall angle sum indentity of tan:

$\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )$

let α+β be t and transform:

$\rm\displaystyle \tan( \gamma ) = - \tan( t)$

remember that tan(t)=tan(t±kπ) so

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )$

therefore when k is 1 we obtain:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )$

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )$

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

$\rm\displaystyle \gamma = - \alpha - \beta \pm \pi$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }$

when is 0:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )$

likewise by Opposite Angle Identity we obtain:

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )$

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

$\rm\displaystyle \gamma = - \alpha - \beta \pm 0$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }$

and we're done!

8 0

3 years ago

Read 2 more answers