Solve this problem. A number c decreased by 9 is less than or equal to 20. Find c.

Mathematics

1 answer:

Lady_Fox [76]3 years ago

5 0

So c-9 is less than or equal to 20 so c is less than or equal to 29. So the correct answer is c. Hope this helps :)

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15 points. Please answer as soon as possible.

abruzzese [7]

Answer:

A. $\frac{(a-b)^{2} +b^{2} }{a^{2}}$

Step-by-step explanation:

So to get the area of a square, we need to find the length of one side.

We know the length of the larger square is a, so the area of the larger cube is $a^{2}$

We can find the length of a side of the smaller square by using pythagoreans theorem to find the hypotenuse of the triangle formed in the bottom left corner. The length of one side along the x axis is a - b, and the length of the other side, along the y-axis, is b.

We can plug it into pythagoreans theorem to get

$(a-b)^{2} +b^{2} = C^{2}$ (C represents the length of one side of the smaller square, and the hypotenuse of the triangle)

$C = \sqrt{ (a-b)^{2} +b^{2}}$

The area of the smaller triangle is C squared to the area of the smaller triangle is

$(a-b)^{2} +b^{2}$

To get the ratio of the smaller square in comparison to the larger square we divide the area of the smaller square by the area of the larger square.

So the ratio should be

$\frac{(a-b)^{2} +b^{2} }{a^{2}}$

8 0

3 years ago

Remember to show work and explain. Use the math font.

vichka [17]

Answer:

$\large\boxed{1.\ f^{-1}(x)=\sqrt[12]{3^x}}\\\\\boxed{2.\ f^{-1}(x)=\sqrt[4]{3^x}}\\\\\ \boxed{3.\ f^{-1}(x)=\sqrt[3]{4^{7-x}}}$

Step-by-step explanation:

$(a^n)^m=a^{nm}\\\\\log_ab=c\iff a^c=b\\\\a^{\log_ax}=x\\\\n\log_ab=\log_ab^n\\\\\log_ab+\log_ac=\log_a(bc)\\============================\\\\1.\\y=3\log_3x^4\to y=\log_3(x^4)^3\to y=\log_3x^{12}$

$2.\\y=\log_3x^4\\\\\text{Exchange x and y. Solve for y:}\\\\\log_3y^4=x\Rightarrow3^{\log_3y^4}=3^x\Rightarrow y^{4}=3^x\\\\y=\sqrt[4]{3^x}\\-------------------------$

$3.\\y=-\log_4x^3+7\\\\\text{Exchange x and y. Solve for y:}\\\\-\log_4y^3+7=x\qquad\text{subtract 7 from both sides}\\\\-\log_4 y^3=x-7\qquad\text{change the signs}\\\\\log_4y^3=7-x\Rightarrow4^{\log_4y^3}=4^{7-x}\\\\y^3=4^{7-x}\Rightarrow y=\sqrt[3]{4^{7-x}}$