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ira [324]
4 years ago
9

What are the solutions to the following system of equations?

Mathematics
1 answer:
Cloud [144]4 years ago
4 0
<span>(2, 4) and (−8, 84)
see the graph</span>

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How to solve the system of 4x+2y-5z=47
MrRa [10]
The answer is x=47/4-y/2+5z/4
6 0
4 years ago
Read 2 more answers
Drag and drop the correct numbers into the boxes to show the initial value and the range for each of the linear functions f(x) a
tatuchka [14]
I think  the answer is the 4th one
6 0
4 years ago
The entire graph of the function f is shown in the figure below. Write the domain and range of f using interval notation?
Arlecino [84]
<h3>Domain:   (-5, 3]</h3><h3>Range:  [-4, 5)</h3>

========================================================

Explanation:

The domain is the set of allowed x values. In terms of a graph, we look at the left most point to see that x = -5 is the smallest x value possible. However, there's an open hole at this endpoint, so -5 itself is actually not part of the domain. So x must be larger than -5. At the same time, x can be as large as x = 3. Look at the very right tip of the graph to find this x value.

So x spans from -5 to 3, excluding -5 but including 3. We would write -5 < x \le 3 which converts to the interval notation (-5, 3]. Note the mix of curved parenthesis and square bracket. The curved parenthesis means to exclude the endpoint, while the square bracket means include the endpoint.

-----------

The range is the set of possible y outputs. Find out the lowest point of the graph. That is when y = -4 and this value is included due to the filled in circle at the endpoint. But we do not include the largest y value y = 5 as there's an open hole at this endpoint.

So the range is the set of y values such that -4 \le y < 5 which in interval notation would be written as [-4, 5)

-----------

So in short, you're looking for the min and max of x and y to get the domain and range respectively. Be sure to exclude any values where there are open holes as those do not count as part of the graph.

7 0
3 years ago
a jogger runs 4 miles per hour faster downhill than uphill. if the jogger can run 5 miles downhill in the same time that it take
Serga [27]

Answer: the rate uphill is 6 mph.

The rate downhill is 10 mph

Step-by-step explanation:

Let x represent the rate at which the jogger ran uphill.

The jogger runs 4 miles per hour faster downhill than uphill. This means that speed at which the jogger ran downhill is (x + 4) mph

Time = distance/speed

if the jogger can runs 5 miles downhill, then the time taken to run downhill is

5/(x + 4)

At the same time, the jogger runs 3 miles uphill. It means that the time taken to run uphill is

3/x

Since the time is the same, it means that

5/(x + 4) = 3/x

Cross multiplying, it becomes

5 × x = 3(x + 4)

5x = 3x + 12

5x - 3x = 12

2x = 12

x = 12/2

x = 6

The rate downhill is 6 + 4 = 10 mph

8 0
4 years ago
Read 2 more answers
Can i pls get some help here with how to even solve it using clear steps? ​
kirill [66]

first off, let's split the triplet into two equations, then from there on we'll do substitution.

\cfrac{y}{x-z}=\cfrac{x}{y}=\cfrac{x+y}{z}\implies \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=\cfrac{x+y}{z} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{\cfrac{y}{x-z}=\cfrac{x}{y}\implies }y^2=\underline{x^2-xz} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 2nd equation}}{\cfrac{x}{y}=\cfrac{x+y}{z}\implies }xz=xy+y^2\implies \stackrel{\textit{substituting for }y^2}{xz=xy+(\underline{x^2-xz})}

2xz=xy+x^2\implies 2xz=x(y+x)\implies \cfrac{2xz}{x}=y+x \\\\\\ 2z=y+x\implies 2=\cfrac{y+x}{z}\implies 2=\cfrac{x+y}{z} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{}{ \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=\cfrac{x+y}{z} \end{cases}}\implies \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=2 \end{cases}\implies \begin{cases} \cfrac{y}{x-z}=2\\[2em] \cfrac{x}{y}=2 \end{cases}

that of course, is only true if x + y, or our numerator doesn't turn into 0, if it does then our fraction becomes 0 and our equation goes south.  Keeping in mind that x,y and z are numeric values that correlate like so.

4 0
2 years ago
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