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3 years ago

HCFof 506 and 1863

Mathematics

1 answer:

Bogdan [553]3 years ago

4 0

Answer:

The hcf of 506 and 1863 is 23

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Help me please !!!!!!!

never [62]

Answer:

Step-by-step explanation:

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3 years ago

Which expression is equivalent to 4 square root 6 divided by 3 root 2?

saw5 [17]

Answer:

$\sqrt[12]{55296}$ /2

Step-by-step explanation:

$\sqrt[4]{6}$ / $\sqrt[3]{2}$ =1.2422

$\sqrt[12]{27}$ /2=0.66 <-- not matching with the top expression.

$\sqrt[4]{24}$ /2=1.11<--not matching with the top expression.

$\sqrt[12]{55296}$ /2=1.2422<-- matches!!

$\sqrt[12]{177147}$ /3=0.91<-- not matching with the top expression.

5 0

3 years ago

10. What makes an equation a quadratic?

CaHeK987 [17]

Answer:

D. The equation has a variable multiplied by itself. A squared variable.

Step-by-step explanation:

A quadratic equation will always start with a term being squared. For example, $8x^{2}$ + 5x + 7 = 44. Another example would be $4x^{2}$ + 2x - 10 = 12. These are just random numbers, but the equation will always look the same! Hope this helps!

6 0

2 years ago

Read 2 more answers

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 $J=19$ .