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frozen [14]
4 years ago
11

What is two expressions where the solution is 19

Mathematics
1 answer:
GuDViN [60]4 years ago
8 0
<span><span><u>Answer </u>
</span><span>10+9=19
75÷3=19


</span><span><u>Explanations </u>
</span><span>Expression is a mathematical statement that contain variables like a and b combined by operational signs. This sign can be ×,÷,+,-etc
A number like 19 has so many expression, for example we have;
10+9=19

57÷3=19
34-15=19
We can have as many expressions as possible. They are actually infinite. 
</span></span>
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Which equation represents a linear function? (5 points) Equation 1: y3 = 2x + 1 Equation 2: y = 3x + 1 Equation 3: y = 5x2 − 1 E
Anna [14]

Answer:

y = 3x + 1

Step-by-step explanation:

' y = 3x + 1 ' is a equation of a linear function. The equation is in slope-intercept form. All slope intercept-form lines are linear.

An equation for a linear function has no exponents, no variables in fractions, and no square roots.

Hope this helps.

8 0
3 years ago
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Please help!
Nuetrik [128]
Hello there! The formula for finding slope is y2 - y1 / x2 - x1, meaning you subtract the first x and y-coordinates, from the second x and y-coordinates. In this case, the formula would be set up as -2 - 5 / 2 -1, because (2, -2) is the second point and (1, 5) is the first point. The answer is C: 2 - 1.
8 0
4 years ago
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What is the remainder r when the polynomial p(x) is divided by (2x - 1)? is (2x - 1) a factor of p(x)? p(x) = 2x2 + 3x + 1?
garik1379 [7]
The remainder is 3. 

You can get this by performing long division to answer the question. Since there is a remainder, it is not a factor. 
3 0
3 years ago
TWO questions! PLEASE HELP ME!
Mariulka [41]

Answer:

The first one is C I don't know the second one though

Step-by-step explanation:

8 0
3 years ago
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What is the slope-intercept form of linear equations?
lara31 [8.8K]

to get the equation of any straight line, we simply need two points off of it, let's use the points from the picture below.

(\stackrel{x_1}{8}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{12}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{3}}}{\underset{run} {\underset{x_2}{12}-\underset{x_1}{8}}}\implies \cfrac{2}{4}\implies \cfrac{1}{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-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{8}) \\\\\\ y-3=\cfrac{1}{2}x-4\implies y=\cfrac{1}{2}x-1

if we already have the slope, and we can see the y-intercept on the table, then we can simply use the slopel-intercept form and plug both of them in.

\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \begin{array}{|c|ll} \cline{1-1} slope\\ \cline{1-1} \cfrac{1}{2}\\ \cline{1-1} y-intercept&\\\cline{1-1} (0~~,~~-1)\\ \cline{1-1} \end{array}~\hfill y=\cfrac{1}{2}x-1

6 0
2 years ago
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