Answer:
For #5 all you have to do is set down to points in order for it to count. The answer is: (0,2) and (3,6)
Step-by-step explanation:
In order to find where exactly you must plot the points, we plug in what we know. In our equation, the 2 is our y intercept, this means that the point is (0,2) (Side Note: here's another example, (0,6) the 6 would be our y intercept)
From the point (0,2) we use the other value given. The other value given in this case, is our slope. When dealing with slope, we use Rise over Run or Rise/Run. This means that from the 2 on the graph, you rise 3 which would be 5 and you run to the right for 4. This now gets you the points (3,6)
Hope this helps with plotting the other questions! :)
(SIDE NOTE: IF THEY GIVE THE VALUE WITH X AS JUST A NUMBER, YOU CAN PLACE A 1 UNDER IT TO TURN IT INTO A FRACTION. FOR EXAMPLE 3X CAN BE TURNED INTO 3/1 WHICH WOULD THEN BE RISE 3 AND RUN 1.
(sooory about the caps, it's just to get your attention.)
Answer:
y-6=-2(x-8)
Step-by-step explanation:
Just google up a point slope calculator :)
Answer:
A ≈ 158.37km²
Step-by-step explanation:
<u>Using the formulas</u>
A=πr2
d=2r
<u>Solving forA</u>
A = 1
/4πd 2
= 1
/4·π·
14.22 ≈ 158.36769km²
Therefore, the are of the circle in nearest hundredth is A ≈ 158.37km²
Each of these limits correspond to the derivative of some function <em>f(x)</em> at a point <em>x</em> = <em>a</em>.
Recall the limit definition of a function <em>f(x)</em> :
<em>f '(x)</em> = lim [<em>h</em> → 0] ( <em>f(x</em> + <em>h)</em> - <em>f(x) </em>) / <em>h</em>
Then if <em>x</em> = <em>a</em>, we get
<em>f '(a)</em> = lim [<em>h</em> → 0] ( <em>f(a</em> + <em>h)</em> - <em>f(a) </em>) / <em>h</em>
<em />
From here, it's easy to identify what each function and point should be:
(a) <em>f </em>(<em>a</em> + <em>h</em>) = (1 + <em>h</em>)¹ʹ³ → <em>f(x)</em> = <em>x </em>¹ʹ³ and <em>a</em> = 1
(that's a 1/3 in the exponent)
(b) <em>f</em> (<em>a</em> + <em>h</em>) = cos(<em>π</em> + <em>h</em>) → <em>f(x)</em> = cos(<em>x</em>) and <em>a</em> = <em>π</em>
<em />
(c) <em>f</em> (<em>a</em> + <em>h</em>) = 5 (4 + <em>h</em>)⁵ → <em>f(x)</em> = 5<em>x</em> ⁵ and <em>a</em> = 4
(d) <em>f</em> (<em>a</em> + <em>h</em>) = exp(4<em>h</em>) = exp(4 (0 + <em>h</em>)) → <em>f(x)</em> = exp(4<em>x</em>) and <em>a</em> = 0
(where exp(<em>x</em>) = <em>eˣ </em>)
Answer:
Step-by-step explanation:
i wasnt listening to my teachers when they was explaining this stuff sorry