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3 years ago

The table below shows the outputs y for different inputs x:

1 answer:

Nutka1998 [239]3 years ago

3 0

Part A: The table does not represent y as a function of x. We can not input 1 and get 4 and 2 as outputs ( or input 3 and get 12 and 6 as outputs ).
Part B : f (120) is a cost for renting a peddleboat for 120 hours;
f (120) = 20 + 10 * 120 = 20 + 1,200 = 1,220.

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What is this simplified?

lisabon 2012 [21]

Answer:

7/4,

Step-by-step explanation:

it cant be simplified

8 0

2 years ago

Read 2 more answers

Indicate whether the statement is true of false.

yawa3891 [41]

The statement is false, as the system can have no solutions or infinite solutions.

<h3>Is the statement true or false?</h3>

The statement says that a system of linear equations with 3 variables and 3 equations has one solution.

If the variables are x, y, and z, then the system can be written as:

$a_1*x + b_1*y + c_1*z = d_1\\\\a_2*x + b_2*y + c_2*z = d_2\\\\a_3*x + b_3*y + c_3*z = d_3$

Now, the statement is clearly false. Suppose that we have:

$a_1 = a_2 = a_3\\b_1 = b_2 = b_3\\c_1 = c_2 = c_3\\\\d_1 \neq d_2 \neq d_3$

Then we have 3 parallel equations. Parallel equations never do intercept, then this system has no solutions.

Then there are systems of 3 variables with 3 equations where there are no solutions, so the statement is false.

If you want to learn more about systems of equations:

brainly.com/question/13729904

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3 0

2 years ago

You have one type of nut that sells for $2.80/lb and another type of nut that sells for $9.60/lb. You would like to have 20.4 lb

max2010maxim [7]

Answer:11.396Ibs of nuts that cost $9.60/Ib and 9.014Ibs that cost $2.80/Ib

Step-by-step explanation:

First we find the cost of the supposed mixture we are to get by selling it $6.60/Ib which weighs 20.41Ibs

Which is 6.6 x 20.41 = $134.64

Now we label the amount of mixture we want to get with x and y

x = amount of nuts that cost $2.8/Ib

y = amount of nuts that cost $9.6/Ib

Now we know the amount of mixture needed is 20.41Ibs

So x + y = 20.41Ibs

And then since the price of the mixture to be gotten overall is $134.64

We develop an equation with x and y for that same amount

We know the first type of nut is $2.8/Ib

So for x amount we have 2.8x

For the second type of nut that is $9.6/Ib

For y amount we have 9.6y

So adding these to equate to $134.64

2.8x + 9.6y = 134.64

So we have two simultaneous equations

x + y = 20.41 (1)

2.8x + 9.6y = 57.148 (2)

We can solve either using elimination or factorization method

I'm using elimination method

Multiplying the first equation by 2.8 so that the coefficient of x for both equations will be the same

2.8x + 2.8y = 57.148

2.8x + 9.6y = 134.64

Subtracting both equations

-6.8y = -77.492

Dividing both sides by -6.8

y = -77.492/-6.8 =11.396

y = 11.396Ibs which is the amount of nuts that cost $9.6/Ib

Putting y = 11.396 in (1)

x + y = 20.41 (1)

x +11.396 = 20.41

Subtract 11.396 from both sides

x +11.396-11.396 = 20.41-11.396

x = 9.014Ibs which is the amount of nuts that cost $2.8/Ibs

8 0

3 years ago

Solve for X kinda confused i don’t know what to do!!

sweet [91]

Answer:

16.9265381881

Step-by-step explanation:

Tangent = Opposite/Adjacent.

Thus, tan62 = x/9 --> tan62 = 1.88072646535, multiply by 9 to get x, 1.88072646535* 9 = 16.9265381881

7 0

2 years ago

1. Write an equation in slope-intercept form for a line passing through (-3,3) with a slope of 1.

PIT_PIT [208]

$\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{3})~\hspace{10em} slope = m\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-3=1[x-(-3)] \\\\\\ y-3=1(x+3)\implies y-3=x+3\implies y=x+6$

$\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-2}{1-(-4)}\implies \cfrac{10}{1+4}\implies \cfrac{10}{5}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=2[x-(-4)]\implies y-2=2(x+4) \\\\\\ y-2=2x+8\implies y=2x+10$

8 0

3 years ago