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3 years ago

Area of a rectangular garden is 6138m2, if the length of length of garden is 99m what is the width?

Mathematics

1 answer:

IRISSAK [1]3 years ago

7 0

Answer:

62 m

Step-by-step explanation:

you do the inverse

6138/99=62

i dont know if this is right so plz tell me if its wrong

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2 years ago

Read 2 more answers

Find the gcf 15, 18, 30 <br> a. 450 <br> b. 540 <br> c. 6 <br> d. 3

poizon [28]

Factors of 15 - 1, 3, 5, 15
Factors of 18 - 1, 2, 3, 6, 9, 18
Factors of 30 - 1, 2, 3, 5, 6, 10, 15, 30

Your answer is D

Hope I helped you :-)

6 0

2 years ago

One of our brainliest, Konrad509, made this:

Reil [10]

$\dfrac{B_x \sqrt{74_x}}{1D_x}+J_x51_x=4G3_x$

A=10, B=11, C=12, etc.

$\dfrac{11\cdot x^0\cdot \sqrt{7\cdot x^1+4\cdot x^0}}{1\cdot x^1+13\cdot x^0}+19\cdot x^0\cdot (5\cdot x^1+1\cdot x^0)=4\cdot x^2+16\cdot x^1+3\cdot x^0\\\\\dfrac{11\sqrt{7x+4}}{x+13}+19(5x+1)=4x^2+16x+3\\\\\dfrac{11\sqrt{7x+4}}{x+13}+95x+19=4x^2+16x+3\\\\11\sqrt{7x+4}+95x(x+13)+19(x+13)=(4x^2+16x+3)(x+13)\\\\11\sqrt{7x+4}+95x^2+1235x+19x+247=4x^3+52x^2+16x^2+208x+3x+39\\\\11\sqrt{7x+4}=4x^3-27x^2-1043x-208\\\\121(7x+4)=(4x^3-27x^2-1043x-208)^2$

$121(7x+4)=(4x^3-27x^2-1043x-208)^2\\\\847x+484=16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433888 x + 43264\\\\16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433041 x +42780=0$

Now, the "only" thing that remains to do is solving the above equation.

While making this problem I only made sure it has a solution. I didn't try to solve it myself and I didn't know it will end up with such "convoluted" polynomial. Sorry to everyone who tried to solve it... m(_ _)m

I think the best way to approach it is using the rational root theorem since we know that $x\in\mathbb{N}$ . Moreover we can deduce that $x\geq19$ since there is $J$ and $J=19$ .

After you succesfully solve it, you should get the answer $x=20$ .

7 0

3 years ago