50 miles since 1 inch equals 10 miles and there's 5 inches that means 5x10=50
X + 0.5(x + 1) + 4(x + 2) = -30 (where x is the lowest integer)
x + 0.5x + 0.5 + 4x + 8 = -30
5.5x = -38.5
x = -7
The required integers are -7, -6 and -5 (answer).
Answer:
Step-by-step explanation:
The equation factors as ...
![f(x)=\dfrac{2}{(x-3)(x-9)}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7B2%7D%7B%28x-3%29%28x-9%29%7D)
Values of x that make the factors zero are the locations of the vertical asymptotes: x = 3 and x = 9.
The <em>first</em> series uses a <em>linear</em> function with - 1 as <em>first</em> element and 1 as <em>common</em> difference, then the rule corresponding to the series is y = |- 1 + x|.
The <em>second</em> series uses a <em>linear</em> function with - 3 as <em>second</em> element as 2 as <em>common</em> difference, then the rule corresponding to the series is y = |- 3 + 2 · x|.
<h3>What is the pattern and the function behind a given series?</h3>
In this problem we have two cases of <em>arithmetic</em> series, which are sets of elements generated by a condition in the form of <em>linear</em> function and inside <em>absolute</em> power. <em>Linear</em> <em>functions</em> used in these series are of the form:
y = a + r · x (1)
Where:
- a - Value of the first element of the series.
- r - Common difference between two consecutive numbers of the series.
- x - Index of the element of the series.
The <em>first</em> series uses a <em>linear</em> function with - 1 as <em>first</em> element and 1 as <em>common</em> difference, then the rule corresponding to the series is y = |- 1 + x|.
The <em>second</em> series uses a <em>linear</em> function with - 3 as <em>second</em> element as 2 as <em>common</em> difference, then the rule corresponding to the series is y = |- 3 + 2 · x|.
To learn more on series: brainly.com/question/15415793
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Answer:
give me brainliest
Given a line AB with Point C outside it.
Mark point D on the line AB. Join CD
With D as center, and any radius, draw an arc intersecting AB at E, and CD at F.
With C as center, and same radius as before, draw an arc intersecting CD at G.
Open a compass to length EF
Now,with G as center, and compass opened the same radius as before, draw an arc intersecting the previous arc at H.
Draw a line m passing through C and H.
Thus m is the line parallel to AB, and passing through point C
∴m∣∣AB
Step-by-step explanation: