Find the slope. A 2 B -1/2 C -2 D 1/2

Solve the following using Substitution method 2x – 5y = -13 3x + 4y = 15

Digiron [165]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$2x-5y=-13, \: 3x+4y=15$

Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

$2x-5y=-13$

Add 5y to both sides of the equation.

$2x=5y-13$

Divide both sides by 2.

$x=\frac{1}{2}\left(5y-13\right) \\$

Multiply $\frac{1}{2}\\$ times 5y - 13.

$x=\frac{5}{2}y-\frac{13}{2} \\$

Substitute $\frac{5y-13}{2}\\$ for x in the other equation, 3x + 4y = 15.

$3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15 \\$

Multiply 3 times $\frac{5y-13}{2}\\$ .

$\frac{15}{2}y-\frac{39}{2}+4y=15 \\$

Add $\frac{15y}{2} \\$ to 4y.

$\frac{23}{2}y-\frac{39}{2}=15 \\$

Add $\frac{39}{2}\\$ to both sides of the equation.

$\frac{23}{2}y=\frac{69}{2} \\$

Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

$\large \underline{ \underline{ \sf \: y=3 }}$

Substitute 3 for y in $x=\frac{5}{2}y-\frac{13}{2}\\$ . Because the resulting equation contains only one variable, you can solve for x directly.

$x=\frac{5}{2}\times 3-\frac{13}{2} \\$

Multiply 5/2 times 3.

$x=\frac{15-13}{2} \\$

Add $-\frac{13}{2}\\$ to $\frac{15}{2}\\$ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

$\large\underline{ \underline{ \sf \: x=1 }}$

The system is now solved. The value of x & y will be 1 & 3 respectively.

$\huge\boxed{ \boxed{\bf \: x=1, \: y=3 }}$

8 0

2 years ago

Calculate the average rate of change over the interval 2

Morgarella [4.7K]

Answer:

Average rate of change over the interval 2<= x <= 5:

y = 3x + 5: 3

y = 3x^2 + 1: 21

y = 3^x: 78

Step-by-step explanation:

2<= x <= 5

Average rate of change over the interval 2<= x <= 5:

y = 3x + 5

y(5) = 3(5) + 5 = 20

y(2) = 3(2) + 5 = 11

Average rate of change = (20 - 11)/(5-2) = 9/3 = 3

y = 3x^2 + 1

y(5) = 3(5^2) + 1 = 75 + 1 = 76

y(2) = 3(2^2) + 1= 13

Average rate of change = (76 - 13)/(5-2) = 63/3 = 21

y = 3^x

y(5) = 3^5 = 243

y(2) = 3^2 =9

Average rate of change = (243-9)/(5-2) = 234/3 = 78

8 0

2 years ago

HELP RN PLEASEEE for a quiz

sattari [20]

Answer:

Step-by-step explanation:

( $x_{1}$ , $y_{1}$ )

( $x_{2}$ , $y_{2}$ )

m = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

parallel lines have the same slopes , perpendicular lines have slope which are opposite reciprocals.

~~~~~~~~~~~~~~~

A( 3 , 5 )

B( - 2 , 7 )

$m_{AB}$ = $\frac{7-5}{-2-3}$ = - $\frac{2}{5}$

C( 10 , 5 )

D( 6 , 15 )

$m_{CD}$ = $\frac{15-5}{6-10}$ = - $\frac{5}{2}$

The answer is (A) Neither

8 0

1 year ago

Can somebody plz use the diagram at the bottom to answer question 10 (only if u know how to do this) thx sm!

jonny [76]

Answer:

isosceles: 2 angles are the same and 2 sides are the same. Equilateral have all the same sides and angles. Saclene have all dfferent angles and sides.

11) isosceles

12) equilateral

13) isosceles or scalene unclear

---------{NOTICE}----------

If you have any additional questions the best way to reach me is via discord

user: michaelsaltandpepper#0001

5 0

2 years ago

Read 2 more answers

PLEASE HELP!!! I WILL GIVE 100 POINTS AND BRAINLIEST ANSWER IF YOU JUST PLEASE HELP ME!!! ;(

Citrus2011 [14]

If you add the standard deviation to the average, it is 14.84. Because Marcus's score was lower than that, he will not receive a certificate.
I think this should be right for the reason and the calculation.

8 0

3 years ago