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Luden [163]
3 years ago
13

How do we find x and y intercepts

Mathematics
1 answer:
IRISSAK [1]3 years ago
5 0

Answer:

To determine the x-intercept, we set y equal to zero and solve for x. Similarly, to determine the y-intercept, we set x equal to zero and solve for y. ...

To find the x-intercept, set y = 0 \displaystyle y=0 y=0.

To find the y-intercept, set x = 0 \displaystyle x=0 x=0.

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Protein, like carbohydrates and fat, contains calories. The following table shows different amounts of protein and their related
soldier1979 [14.2K]

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yes the answer is proportional!! :)

Step-by-step explanation:

4 0
3 years ago
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345mi +3hr+ 15 mph the slowest car traveled
Ksju [112]

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15mph

Step-by-step explanation:

it is going slow

7 0
3 years ago
Have a look of the drawing... What is "d" value? I have tried many methods but don't find how to solve... Pleaseeee
Phoenix [80]
I´d say "d" is the distance from the eye to the wall. 
Now substracting 1.2-1 you´ll get the distance of the wall of the smallest triangle = 0.2 And you do 1.5-0.2= 0.3 that´s the distance of the wall of the other triangle. Then you solve everything with Pitagoras theorem. You have 2 rectangle triangles.
B+alfa=45°
tan^-1(0.2/d)=B
tan^-1(1.3/d)=alfa
THEN:
tan^-1(0.2/d)+tan^-1(1.3/d)=45°
 Now you have 3 ecs and 3 variables.
alfa,B and "d"

3 0
3 years ago
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HELP ME PLEASEEEEEEEE
stepladder [879]

Answer:

x = 21

Step-by-step explanation:

3x + 18 + 4x + 15 = 180

7x + 33 = 180

7x = 147

x = 21

8 0
3 years ago
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If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
Alexxx [7]
\bf f(x)=y=2x+sin(x)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\

\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}

now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
\\\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
5 0
3 years ago
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