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BaLLatris [955]
3 years ago
10

Is y = x^2 + 1 a function

Mathematics
1 answer:
Nutka1998 [239]3 years ago
7 0
Yes the graph of this will pass the vertical line test meaning no one value of x will produce two different y values
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X=? <br><br> 38 degrees 75 degrees
Nadya [2.5K]
<h2>sum of angle of triangle is 180°</h2>

<h2>A+B+C = 180 </h2>

<h2>38°+75°+ X = 180°</h2>

<h2>113° + X = 180°</h2>

<h2>X = 180° -113</h2>

<h2>X = 67°</h2>

4 0
3 years ago
Identify the maximum and minimum values of the function y = 8 cos x in the interval [-2π , 2π].
Veronika [31]
Hello here is a solution ; 

3 0
3 years ago
A gold mine has two​ elevators, one for equipment and another for the miners. The equipment elevator descends 4 feet per second.
Ad libitum [116K]

Let d_m,\ d_e represent the depth of the miner and equipments elevator, respectively. We know that equipment elevator descends with a velocity of v_e = 4 feet per second, and the miners one descends with a velocity of v_m = 15 feet per second.

If we start counting time (t=0) when the equipment elevator begins to descend, after t seconds its depth will be

d_e(t) = v_et = 4t

On the other hand, the miner elevator follows the same rule, but we have to use its velocity, and remember that it starts with a delay of thirty seconds:

d_m(t) = v_m(t-30) = 15(t - 30)

Now, we have to wait the 30 seconds of delay, and then another 14 seconds. This means that we want to know the positions of both elevators when t = 44. Let's plug this value into the two equations:

d_e(44) = 4\cdot 44 = 176

d_m(44) = 15(44-30) = 15\cdot 14 = 210

So, the equipment elevator is 176 feet deep, and the miner elevator is 210 feet deep, and thus this is the deeper one.

7 0
3 years ago
The volume
Sedaia [141]
\bf \begin{array}{cccccclllll}&#10;\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\&#10;\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\&#10;y&=&{{ k}}&\cdot&x&#10;&&  y={{ k }}x&#10;\end{array}\\ \quad \\&#10;

and also

\bf \begin{array}{llllll}&#10;\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\&#10;\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\&#10;y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}&#10;&&y=\cfrac{{{  k}}}{x}&#10;\end{array}&#10;

now, we know that V varies directly to T and inversely to P simultaneously
thus\bf V=T\cdot \cfrac{k}{P}

so     \bf V=T\cdot \cfrac{k}{P}\qquad &#10;\begin{cases}&#10;V=42\\&#10;T=84\\&#10;P=8&#10;\end{cases}\implies 42=\cfrac{84k}{8}\implies 4=k&#10;\\\\\\&#10;V=\cfrac{4T}{P}\qquad now\quad &#10;\begin{cases}&#10;V=74\\&#10;P=10&#10;\end{cases}\implies 74=\cfrac{4T}{10}\implies 185=T
7 0
3 years ago
What is the pattern of 1.1, 2.2, 3.3, 4.4
Tanya [424]
It is adding one little percentage of increase here. or, 1.1 each time....hope that this helps.
5 0
3 years ago
Read 2 more answers
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