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3 years ago

Solve the formula V = bh for h

Mathematics

2 answers:

alex41 [277]3 years ago

6 0

Solving for h means to isolate h by move everything else to the other side of the equal sign.

V = b * h

$\frac{V}{b} = \frac{b*h}{b}$ divide both sides by b

$\frac{V}{b} = h$ h is now isolated

mixas84 [53]3 years ago

3 0

V = bh Dividing both sides by "b" we get

(V / b) = h

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insens350 [35]

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2 years ago

Which set of ordered pairs do NOT represent a function?

Rainbow [258]

Answer:

Every input can have only one output. But an output can have multiple inputs. Basically an x coordinate can have the same y coordinate as another ordered pair but not the same x coordinate. This is why d) does not represent a function.

Step-by-step explanation:

(4,-2)

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3 years ago

In two or three sentences explain how you would solve for the real solutions of the following equation: please help asap!!

g100num [7]

the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

Step-by-step explanation:

$x^{3}$ = $- 64$

$x^{3} +64$ = 0

We can write 64 as $4^{3}$

$x^{3}$ + $4^{3}$ = 0

using the identity ( $x^{3}+y^{3} = (x+y)(x^{2} -xy+y^{2} )$ )

we get,

= $(x+4) (x^{2} -x*4+4^{2} )$

= $(x+4)(x^{2} -4x+16)$ ....................(1)

solving the quadratic equation ,

$x^{2} -4x+16$ =0

solutions of this quadratic equation can be obtained by

$x=-b +- \sqrt{b^{2}-4ac } /2a$

let use y for factors

$x=-(-4x)+-\sqrt{(-4x^{2} )-4*x^{2} *16} / 2*x^{2}$

$x=4x+-\sqrt{16x^{2} -64x^{2} } /2x^{2}$

$x=4+-\sqrt{16-64}/2$

$x=4+-4\sqrt{3i} /2$

$x=2+-2\sqrt{3i}$ ..................(2)

from the equation 1 we have,

$x-4=0$

which gives solution $x=4$

and from equation 2 we got $x=2+-2\sqrt{3i}$

so the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

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Answer:

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2 years ago

Read 2 more answers