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3 years ago

What is 16-14 dealing with intergers

2 answers:

5 0

16-14=2

Think of it on the number line. You are at the integer 16. You are subtracting 14 from 16 so you move 14 integers back from 16 on the number line. The resulting integer is 2.

You can also take food for example. If you have 16 chocolates and you eat 14 you will have 2 left.

fenix001 [56]3 years ago

3 0

16 - 14 = 2

Think of it as in objects. You have 16 candies, and you eat 14. How many do you have left? You can count it and see. You have 2 left. So 16 - 14 = 2

Hope that helps :)

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To graph the equation 2x + 5y = 10, Zeplyn draws a line through the points (5, 0) and (0, What is the slope of the line represen

wariber [46]

Answer: -2/5

Step-by-step explanation:

To find the slope, we set the function use y = mx + b forms.

2x + 5y = 10

5y = -2x + 10 Minus 2x in both side

y = -2/5x + 10/5 Divide 5 from both side

y = -2/5x + 2

In the y = mx + b, the m before variable x is always the slope, so -2/5 is the slope for 2x + 5y = 10.

8 0

3 years ago

1. Write an addition equation, a subtraction equation, and a multiplication equation that has -2 as a solution. Solve each equat

Sphinxa [80]

5 + -7 = -2
5 - 7 = -2
-1 x 2 = -2

7 0

3 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$