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3 years ago

Find two consecutive integers whose sum is -29

Mathematics

1 answer:

LuckyWell [14K]3 years ago

8 0

Example:

The sum of two consecutive integers is -15. What is the product of the two integers?

x = 1st consecutive integer

x + 1 = 2nd consecutive integer {consecutive integers increase by 1}

x + x + 1 = -15 {the sum of two consecutive integers is -15}

2x + 1 = -15 {combined like terms}

2x = -16 {subtracted 1 from each side}

x = -8 {divided each side by 2}

x + 1 = -7 {substituted -8, in for x, into x + 1}

-8 and -7 are the two consecutive integers

Their product is 56.

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Please please help me out

masya89 [10]

Answer: y = 3x + 1

Step-by-step explanation:

The equation is in Slope-Intercept format: y = mx + b where

m is the slope (difference in y's over difference in x's)
b is the y-intercept (value of y when x = 0)

$m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\\\text{Let }(x_1,y_1)=(0,1)\quad \text{and}\quad \text{let }(x_2, y_2)=(1,4)\quad then\\\\m=\dfrac{4-1}{1-0}=\dfrac{3}{1}=3$

b = 1 refer to the table that shows y = 1 when x = 0

Now input m = 3 and b = 1 into the Slope-Intercept formula:

y = 3x + 1

8 0

3 years ago

Check whether 2-3x is a factor of (x)2 -x+1

sergejj [24]

Answer:

Step-by-step explanation:

if it was a factor then, 2-3x = 0

x= 2/3

putting x = 2/3 in eqn,

(2/3)^2 - 2/3 +1 =7/9....so it is not a factor

5 0

3 years ago

Read 2 more answers

156872/3756 quotient remainder

andreev551 [17]

41.8 rounded to the nearest tenth

3 0

3 years ago

Look at the graph shown below:

ElenaW [278]

So.. take a peek at the picture... let's get two points from it, hmm say 0,4 notice it touches the y-axis there, and say hmmm -4, 1, almost at the bottom of the line

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 0}}\quad ,&{{ 4}})\quad 
% (c,d)
&({{ -4}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-4}{-4-0}$

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad 
\begin{array}{llll}
\textit{plug in the values for }
\begin{cases}
y_1=4\\
x_1=0\\
m=\boxed{?}
\end{cases}\\
\textit{and solve for "y"}
\end{array}\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}$

once you get the slope and solve for "y", that'd be the equation of the line.

7 0

3 years ago

Can Anyone Help Me

klio [65]

Candy bar, Bicycle, 12-Story Apartment Building, Pick-Up Truck.

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3 years ago