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Lostsunrise [7]

1 year ago

12

. Write a number in which the value of the 6 is one tenth the value of the 6 ir 35,629.

1 answer:

Semenov [28]1 year ago

4 0

Answer:

65

Step-by-step explanation:

You might be interested in

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

Elan Coil [88]

Answer:

Domain: $-3\leq x\leq 1$

Range: $-4\leq y\leq 0$

Step-by-step explanation:

If you find the coordinate of the graph you can get the domain and range

(-3,0), (-2,-4), (-1,-2), (0,0), (1, -4)

Now that we have that find the domain and range

Domain - x coordinate

Range - y coordinate

Domain: (-3, -2, -1, 0, 1)

Range: (-4, -2, 0)

Since the domain ranges from -3 to 1 you can use the inequality $-3\leq x\leq 1$ to represent the domain and $-4\leq y\leq 0$ for the range

7 0

2 years ago

Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.

zloy xaker [14]

Answer:

$y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

$x={\pi}{4}$

Step-by-step explanation:

We are given that

$y'=\frac{2cos2x}{3+2y}$

y(0)=-1

$\frac{dy}{dx}=\frac{2cos2x}{3+2y}$

$(3+2y)dy=2cos2x dx$

Taking integration on both sides then we get

$\int (3+2y)dy=2\int cos 2xdx$

$3y+y^2=sin2x+C$

Using formula

$\int x^n=\frac{x^{n+1}}{n+1}+C$

$\int cosx dx=sinx$

Substitute x=0 and y=-1

$-3+1=sin0+C$

$-2=C$

$sin0=0$

Substitute the value of C

$y^2+3y=sin2x-2$

$y^2+3y-sin 2x+2=0$

$y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

Hence, the solution $y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

When the solution is maximum then y'=0

$\frac{2cos2x}{3+2y}=0$

$2cos2x=0$

$cos2x=0$

$cos2x=cos\frac{\pi}{2}$

$cos\frac{\pi}{2}=0$

$2x=\frac{\pi}{2}$

$x=\frac{\pi}{4}$

3 0

2 years ago

Is 2/3 greater than 10/18

SVEN [57.7K]

2/3 is greater than 10/18

4 0

2 years ago

Read 2 more answers

Please help super easy. (am dum)

ycow [4]

Answer:

Here is the answer

Step-by-step explanation:

36 = 3x2x3x2

8 = 2x2x2

42 = 7x3x2

4 0

2 years ago