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3 years ago

What is the value of the function when x = 0? x −2 0 1 3 6 y 1 3 4 2 0

1 answer:

6 0

$\begin{array}{ccccccccccc}x&|&-2&|&\boxed0&|&1&|&3&|&6\\-&-&-&-&-&-&-&-&-&-&-\\y&|&1&|&\boxed3&|&4&|&2&|&0\end{array}\\.\\\\\boxed{f(0)=3}$

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-11y = 6(2+1) - 13y

iris [78.8K]

Answer:

Step-by-step explanation:

Let's put the equations in standard form. For the first equation, we have:

−11y=6(z+1)-13y

2y−6z=6

y−3z=3

The second equation is:

4y−24=c(z−1)

4y−cz=24−c

If we multiply the first equation by 4, we get:

4y-12z=12

Comparing the two equations, we see that if c=12, both equations will be the same and there will be infinitely many solutions.

The correct value of c is 12.

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3 years ago

What is a non-example of a independent variable?

alexandr402 [8]

An non-example of a independent variable is how much money you make selling cookies , because it depends on the number of cookies you sell .

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3 years ago

Read 2 more answers

500m:2km in it's lowest form<br>

blagie [28]

500m:2km (change km to m)

500m:2000m (divide by 100)

5m:20m (divide by 5)

1m:4m

7 0

3 years ago

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

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3 years ago

What type of slope does

sukhopar [10]

Answer:

Step-by-step explanation:

The equation is in slope-intercept form, so the slope is the coefficient of x. Since the coefficient of x is negative, so is the slope.

4 0

1 year ago