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True [87]
3 years ago
8

Plz tell me the answer!!!!

Mathematics
1 answer:
Anestetic [448]3 years ago
7 0
Monday and Wednesday since they both round to 1.45
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Write an equation that represents the function. (linear equations)
irinina [24]

Answer:

y = 2x + 2

Step-by-step explanation:

First solve for the slope using \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

\frac{8-2}{3-0} = 2

Plug the slope into the slope intercept form equation y = mx + b

y = 2x + b

To solve for b, plug in any (x, y) point on the table.

2 = 2(0) + b

2 = b

Now you can plug the b value into your equation.

y = 2x + 2

3 0
2 years ago
What is the domain of the function f(x)= e^x/e^x+c if c is a constant greater than 0
Viktor [21]

Answer: Option d.

Step-by-step explanation:

To find the domain of the function we should look for the values for which the denominator is equal to zero, because the division by zero is not allowed.

We know by definition that the function e^x is always greater than zero for all <em>x</em>.

We know that the constant <em>c</em> is greater than zero (c>0).

Then, the expression e^x+c is never equal to zero.

Therefore, it does not exist a value for <em>x</em> that makes the denominator 0. Then, the domain of the function is all real numbers.

The answer is the option d.

3 0
3 years ago
Special right triangles <br><br> Find the value of each missing variable
Shalnov [3]

Hi!

Rules for solving 30-60-90 triangles are as follows:

* The hypotenuse is double the short leg.

* The short leg times √3 is the long leg

To solve then, we divide 4√21 by √3 and get 4√7. That is x, and to get y we multiply by 2. That is 8√7.

Hope this helps! :D

6 0
3 years ago
Match each graph with its related table. Explain your answers.
Oksanka [162]

Answer:

6. = A.

5. = B.

4. = C.

Step-by-step explanation:

5 0
3 years ago
Find the matrix product, if possible.  Show your work please
steposvetlana [31]
\left[\begin{array}{ccc}-1&3\\5&4\end{array}\right] \cdot  \left[\begin{array}{ccc}0&-2&6\\1&-3&2\end{array}\right]\\\\=  \left[\begin{array}{ccc}(-1)(0)+(3)(1)&(-1)(-2)+(3)(-3)&(-1)(6)+(3)(2)\\(5)(0)+(4)(1)&(5)(-2)+(4)(-3)&(5)(6)+(4)(2)\end{array}\right]\\\\=  \left[\begin{array}{ccc}0+3&2-9&-6+6\\0+4&-10-12&30+8\end{array}\right]\\\\  =\left[\begin{array}{ccc}3&-7&0\\4&-22&38\end{array}\right]

3 0
3 years ago
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