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3 years ago

Write the equation of the line in slope-intercept form with a slope of -5 and a y-intercept of -1.

2 answers:

7 0

First, we need to know what is the general equation of the slope-intercept form. It is expressed as follows:

y = mx + b

where m is the slope and b is the y-intercept. Therefore, having a slope of -5 and a y-intercept of -1, the equation of the line should be:

y = -5x - 1

Ghella [55]3 years ago

6 0

Slope intercept form : y = mx + b

the slope is represented by m....so m = -5
the y intercept is represented by b....so b = -1

now we sub
y = mx + b
y = -5x + (-1) or y = -5x - 1 <===

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Which ordered pairs in the form (x, y)(x, y) are solutions to the equation

kap26 [50]

The answer is (-1,-7) because you are pluging it in -1 is x and -7 is y. so 7×-1=-7 and 5×-7 = -35 so -7 minus -35 =28

6 0

3 years ago

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

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krek1111 [17]

7.065 grams of tin I believe

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2 years ago

The expression 35 exponet 2 is equivalent to the product of two perfect squares (each greater than 1). You can find the two

raketka [301]

Answer:

Step-by-step explanation:

7 0

3 years ago

Figure ABCD has verticies A(-4, 1) B(2, 1) C(2, -5) D(-4-3). What was the area of Figure ABCD.

JulijaS [17]

area: 18 units^2

Step-by-step explanation:

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8 0

2 years ago

Read 2 more answers