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vova2212 [387]
3 years ago
7

Determine the intercept of the line

Mathematics
1 answer:
Vadim26 [7]3 years ago
6 0
Y intercept: when x = 0
Look on the graph
It would be (0, 0.4)

X intercept: when y = 0
Look on the graph
It would be (0.3, 0)
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Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!
valkas [14]

Answer by JKismyhusbandbae: expression 2 and expression 1

Look at the four expressions. Simplify any expressions that can be simplified to see which two are equivalent.

8v × 30v = ( 8 × 30) × ( v × v) = 240v^2

Since expression 2 can be simplified to expression 1, they are equivalent.

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3 years ago
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B(x)=x2+3
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3 years ago
I don't know if this is right... please someone help mee
worty [1.4K]
For the first circle, let's use the pythagorean theorem

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

 so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.



now, let's check for second circle

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938

well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.



let's check the third circle

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
6 0
2 years ago
Fractions, Decimals, and Percents
KatRina [158]

Answer:

Step-by-step explanation:

13.84%,0.84

14.10,12,12,14,15,17,20.Median is 14.Mode is 12.

15.Mean so add 18+19+20=57.57/3 =19.

8 0
3 years ago
Graph the image of H(-8,5) after a reflection over the x-axis.<br><br> Answer ?
TEA [102]

Answer: plot a point at (-8, -5)

The y coordinate flips from positive to negative, or vice versa, when we reflect over the horizontal x axis. The x coordinate stays the same.

The rule can be written as (x,y) \to (x,-y)

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3 years ago
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