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3 years ago

Work out the area of a triangle of bass 12 cm and height 7 cm

Mathematics

2 answers:

ICE Princess25 [194]3 years ago

8 0

1/2 (b x h) 1/2(12) x7 6 x 7 **42**

Paul [167]3 years ago

7 0

The formula for the area of a triangle is 1/2*b*h
The best thing to do is to multiply the answers then half them.
7*12= 84
84/2= 42
Therefore, the area is 42cm2
Hope this helps :)

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The local humane society has 5 cats, 9 dogs, and 2rabbits that need home. What fractional part of these pets are cats?

kykrilka [37]

Answer:

5/16 i think

Step-by-step explanation:

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3 years ago

Sales increased by only 12% last month. If the sales from the previous month were $152,850, what were last months sales

Tamiku [17]

Answer:

$\$171,192$

Step-by-step explanation:

Remember that

$100\%+12\%=112\%=112/100=1.12$

we know that

The sales from the previous month were $152,850

This amount represent the 100%

Find out what were last months sales, multiply the sales from the previous month by the factor 1.12

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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

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

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

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

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

<u /> $x=2+-2\sqrt{3i}$ ..................(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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