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Yuliya22 [10]
2 years ago
9

The matrix to solve

Mathematics
2 answers:
luda_lava [24]2 years ago
8 0
4x-4=0
8x+4=8

4x=4
8x=4

4x=8x

It doesn't work so this is proof that this is incorrect.

Hope this helps :)
Maurinko [17]2 years ago
4 0
This question is impossible to solve.

this is why:

if 4x-4=0
then x=1

so if 8x + 4 = 8

then x is false.

I get these question quite a lot in my GCSEs.

you have to prove why they are fake.

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This question is down below. The picture attached below. <br> Thanks.
Ostrovityanka [42]

The vendor has to sell 88 gingerbread houses to earn a profit of $665.60 and there is no chance that the vendor will earn $1500.

Given an equation showing profits of A Christmas vendor as

P=-0.1g^{2}+30g-1200.

We have to find the number of gingerbread houses that the vendor needs to sell in order to earn profit of $665.60 and $1500.

To find the number of gingerbread houses we have to put P=665.60 in the equation given which shows the profit earned by vendor.

665.60=-0.1g^{2}+30g-1200

0.1g^{2}-30g+1200+665.60=0

0.1g^{2}-30g+1865.60=0

Divide the above equation by 0.1.

g^{2}-300g+18656=0

Solving for g we get,

g=[300±\sqrt{(300)^{2}-4*1*18656 }]/2*1

g=[300±\sqrt{90000-74624}]/2

g=[300±\sqrt{15376}]/2

g=(300±124)/2

g=(300+124)/2       , g=(300-124)/2

g=424/2,  g=176/2

g=212,88

Because 212 is much greater than 88 so vendor prefers to choose selling of 88 gingerbread houses.

Put the value of P=1500 in equation P=-0.1g^{2}+30g-1200.

-0.1g^{2}+30g-1200=1500

0.1g^{2}-30g+1500+1200=0

0.1g^{2}-30g+2700=0

Dividing equation by 0.1.

g^{2}-300g+27000=0

Solving the equation for finding value of g.

g=[300±\sqrt{300^{2} -4*1*27000}]/2*1

=[300±\sqrt{90000-108000}] /2

=[300±\sqrt{-18000}]/2

Because \sqrt{-18000} comes out with an imaginary number so it cannot be solved for the number of gingerbread houses.

Hence the vendor has to sell 88 gingerbread houses to earn a profit of $665.60 and there is no chance that the vendor will earn $1500.

Learn more about equation at brainly.com/question/2972832

#SPJ1

7 0
1 year ago
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