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Virty [35]
3 years ago
10

Find the constant of proportionality for the table and write in the form y = kx.

Mathematics
2 answers:
Elden [556K]3 years ago
5 0

Answer:

it b

Step-by-step explanation:

4vir4ik [10]3 years ago
4 0

Answer:

Y=4X

Step-by-step explanation:

Hope this helps

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Find the GCF (greatest common factor) of the following terms.<br> {5x2y2,25x3y2,x2y2}
HACTEHA [7]

Answer:

5x2y2 is the GCF common factor of the following terms.

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2 years ago
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Which expression is equivalent
Tatiana [17]

Given expression is

\sqrt[4]{\frac{16x^{11}y^8}{81x^7y^6}}

Radical is fourth root

first we simplify the terms inside the radical

\frac{x^{11}}{x^7}=x^4

\frac{y^(8)}{y^6}=y^2

So the expression becomes

\sqrt[4]{\frac{16x^4y^2}{81}}

Now we take fourth root

\sqrt[4]{16} = 2

\sqrt[4]{81} = 3

\sqrt[4]{x^4} = x

We cannot simplify fourth root (y^2)

After simplification , expression becomes

\frac{2x\sqrt[4]{y^2}}{3}

Answer is option B


8 0
3 years ago
What is f(g(x)) for x &gt; 5?
mel-nik [20]

Answer:

\large\boxed{B.\ 4x^2-41x+105}

Step-by-step explanation:

f(x)=4x-\sqrt{x}\\\\g(x)=(x-5)^2\\\\f(g(x))\to\text{put}\ x=(x-5)^2\ \text{to}\ f(x):\\\\f(g(x))=f\bigg((x-5)^2\bigg)=4(x-5)^2-\sqrt{(x-5)^2}\\\\\text{use}\\(a-b)^2=a^2-2ab+b^2\\\sqrt{x^2}=|x|\\\\f(g(x)=4(x^2-2(x)(5)+5^2)-|x-5|\\\\x>5,\ \text{therefore}\ x-5>0\to|x-5|=x-5\\\\f(g(x))=4(x^2-10x+25)-(x-5)\\\\\text{use the distributive property:}\ a(b+c)=ab+ac\\\\f(g(x))=(4)(x^2)+(4)(-10x)+(4)(25)-x-(-5)\\\\f(g(x))=4x^2-40x+100-x+5\\\\\text{combine like terms}\\\\f(g(x))=4x^2+(-40x-x)+(100+5)\\\\f(g(x))=4x^2-41x+105

6 0
3 years ago
Read 2 more answers
On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th
Elena L [17]

Answer:

The area of triangle for the given coordinates is  1.5\sqrt{4.6}

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, a = \sqrt{(3-0)^{2}+(4-0)^{2}}

Or, a =  \sqrt{9+16}

Or, a =   \sqrt{25}

∴ a = 5 unit

Similarly

The measure of length BC = b = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, b = \sqrt{(3-3)^{2}+(2-4)^{2}}

Or, a =  \sqrt{0+4}

Or, b =   \sqrt{4}

∴ b = 2 unit

And

So, The measure of length CA = c = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, c = \sqrt{(3-0)^{2}+(2-0)^{2}}

Or, c =  \sqrt{9+4}

Or, c =   \sqrt{13}

∴ c = \sqrt{13} unit

Now, area of Triangle written as , from Heron's formula

A = \sqrt{s\times (s-a)\times (s-b)\times (s-c)}

and s = \frac{a+b+c}{2}

I.e  s = \frac{5+2+\sqrt{13}}{2}

Or. s =  \frac{7+\sqrt{13}}{2}

So, A = \sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}

Or, A = \sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}

Or, A = \frac{3}{2} × \sqrt{1+\sqrt{13} }

∴  Area of triangle = 1.5\sqrt{4.6}

Hence The area of triangle for the given coordinates is  1.5\sqrt{4.6}  Answer

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3 years ago
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