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3 years ago

Which is the only center point that lies on the edge of a triangle?

Mathematics

2 answers:

ELEN [110]3 years ago

5 0

Answer:

The Circumcenter of a right triangle.

Step-by-step explanation:

Rus_ich [418]3 years ago

4 0

The circumcenter is also center of the triangles circumcircle. i'm not 100% but if this is one of your answers then i hoped this helped!

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A class voted for either rollerblading, swimming or biking as their favorite summer activity. If rollerblading got 10% percent o

weeeeeb [17]

The answer would be 55%

6 0

2 years ago

What the y-intercept of (8,0) and (3,7)

Anettt [7]

Answer:

$\sf \bf y - intercept = \dfrac{56}{5}$

Step-by-step explanation:

Y-intercept:

(8,0) ; x₁ = 8 & y₁ = 0

(3,7) ; x₂= 3 & y₂ = 7

First find the slope of the line.

$\sf \boxed{\bf Slope =\dfrac{y_2-y_1}{x_2-x_1}}$

$\sf = \dfrac{7 - 0}{3 - 8}\\\\ =\dfrac{7}{-5}\\\\=\dfrac{-7}{5}$

slope y-intercept form of the line: y = mx + b

Substitute the slope and x and y in the above equation. We can choose anyone of the given points.

$0 = \dfrac{-7}{5}*8 + b\\\\0=\dfrac{-56}{5}+b\\\\\dfrac{56}{5}=b\\\\ \boxed{\bf b = \dfrac{56}{5}}$

3 0

1 year ago

If anyone knows this can you let me know please. :) thank you. No scams please

Genrish500 [490]

Answer:

h(0)= -7

Step-by-step explanation:

that is my answer

3 0

3 years ago

Solve the following problem, using mixed number format.<br><br> 2 and 1/2 + 1 and 1/16 =

77julia77 [94]

Step-by-step explanation:

2 1/2 + 1 1/16

Adding whole number parts and fraction parts together,

(2+1) + (1/2 + 1/16)

= 3 + (1*8/2*8 + 1/16)

= 3 + (8/16 + 1/16)

= 3 + 9/16

= 3 and 9/16

6 0

2 years ago

F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))

Marrrta [24]

Given:

$f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6$

Required:

$We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.$

Explanation:

Given equation is

$F(x)=\sqrt[4]{f(x)g(x)}.$ $F(x)=(f(x)g(x))^{\frac{1}{4}}$ $F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}$

Differentiate the given equation for x.

$Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{ Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.$

$F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)$ $F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)$

Replace x=3 in the equation.

$F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)$ $Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}$ $F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)$ $F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-6-1}{4}$ $F^{\prime}(3)=\frac{-7}{4}$

Final answer:

$F^{\prime}(3)=\frac{-7}{4}$

8 0

8 months ago