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3 years ago

Which function passes through the points (2, 15) and (3, 26)?

Mathematics

2 answers:

Alik [6]3 years ago

6 0

Answer:

the equation is y = 11x - 7

B is correct option.

Step-by-step explanation:

The function passes through the points (2, 15) and (3, 26)

Slope can be calculated by the formula

$m=\frac{y_2-y_1}{x_2-x_1}$

using this formula, the slope is given by

$m=\frac{26-15}{3-2}\\\\m=\frac{11}{1}\\\\m=11$

The slope intercept form of line is y = mx+b

here, m = 11

hence, the equation is y = 11x +b

Now, using the point (2,15) to find b

15=11(2)+b

15 = 22 +b

b = -7

Hence, the equation is y = 11x - 7

B is correct option.

Blizzard [7]3 years ago

4 0

Answer:

Step-by-step explanation:

We can solve this by finding the slope of the function that passes through the points (2,15) and (3,26). We can use the "formula" rise over run.

So we have:

(26-15)/(3-2) which gives us 11 as our slope. Now we must find the y intercept!

It is -7.

So the answer is B

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Mashutka [201]

11x+9

Step by step:
Remove the parenthesis
3x+7+8x+2
Collect like terms
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2 years ago

Choose the equation that satisfies the data in the table.

Zanzabum

Answer:

B. y = $\frac{2}{15} X + 2$

Step-by-step explanation:

I. Given x = 15 and y = 4

4 = $\frac{2}{15}(15) + 2$

4 = 2+2

4 = 4 true

II. Give x = 0 and y = 2

2 = $\frac{2}{15}(0) + 2$

2 = 0 + 2

2 = 2 true

III. Given x = -15 and y = 0

0 = $\frac{2}{15}(-15) + 2$

0 = -2 + 2

0 = 0 true

Good luck.

6 0

3 years ago

Carlos is using the quadratic formula to find the solutions of y=3x^2-5x-2. Which of the following will simplify to the correct

Nana76 [90]

Answer:

$Fx=\frac{5\pm\sqrt{25+24} }{6}$

Step-by-step explanation:

A quadratic equation in one variable given by the general expression:

$ax^2+bx+c$

Where:

$a\neq 0$

The roots of this equation can be found using the quadratic formula, which is given by:

$x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

So:

$y(x)=3x^2-5x-2=0$

As you can see, in this case:

$a=3\\b=-5\\c=-2$

Using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}=\frac{-(-5)\pm\sqrt{(-5)^2-4(3)(-2)} }{2(3)}=\frac{5\pm\sqrt{25+24} }{6}$

Therefore, the answer is:

$Fx=\frac{5\pm\sqrt{25+24} }{6}$

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