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2 years ago

6

The work of a student to solve a set of equations is shown: Equation A: y = 4 − 2z Equation B: 4y = 2 − 4z Step 1: −4(y) = −4(4

− 2z) [Equation A is multiplied by −4.] 4y = 2 − 4z [Equation B] Step 2: −4y = 4 − 2z [Equation A in Step 1 is simplified.] 4y = 2 − 4z [Equation B] Step 3: 0 = 6 − 6z [Equations in Step 2 are added.] Step 4: 6z = 6 Step 5: z = 1 In which step did the student first make an error? (5 points)

2 answers:

mina [271]2 years ago

8 0

Answer:

step 2.

Step-by-step explanation:

dem82 [27]2 years ago

4 0

Answer:

The student made the first mistake in step 2.

Step-by-step explanation:

You might be interested in

1. Which point on the axis satisfies the inequality y

grigory [225]

Answer:

1) Point $(1,0)$ -----> see the attached figure N $1$

2) The value of x is $4$

3) I quadrant

4) $(1,1)$

5) $y>-5x+3$

Step-by-step explanation:

Part 1)

we know that

If the point satisfy the inequality

then

the point must be included in the shaded area

The point $(1,0)$ is included in the shaded area

Part 2)

we have

$x-2y\geq 4$

see the attached figure N $2$

we know that

The value for x on the boundary line and the x axis is equal to the x-intercept of the line $x-2y= 4$

For $y=0$

Find the value of x

$x-2(0)= 4$

$x=4$

The solution is $x=4$

Part 3)

we have

$x\geq 0$ -----> inequality A

The solution of the inequality A is in the first and fourth quadrant

$y\geq 0$ -----> inequality B

The solution of the inequality B is in the first and second quadrant

so

the solution of the inequality A and the inequality B is the first quadrant

Part 4) Which ordered pair is a solution of the inequality?

we have

$y\geq 4x-5$

we know that

If a ordered pair is a solution of the inequality

then

the ordered pair must be satisfy the inequality

we're going to verify all the cases

<u>case A)</u> point $(3,4)$

Substitute the value of x and y in the inequality

$x=3,y=4$

$4\geq 4(3)-5$

$4\geq 7$ ------> is not true

therefore

the point $(3,4)$ is not a solution of the inequality

<u>case B)</u> point $(2,1)$

Substitute the value of x and y in the inequality

$x=2,y=1$

$1\geq 4(2)-5$

$1\geq 3$ ------> is not true

therefore

the point $(2,1)$ is not a solution of the inequality

<u>case C)</u> point $(3,0)$

Substitute the value of x and y in the inequality

$x=3,y=0$

$0\geq 4(3)-5$

$0\geq 7$ ------> is not true

therefore

the point $(3,0)$ is not a solution of the inequality

<u>case D)</u> point $(1,1)$

Substitute the value of x and y in the inequality

$x=1,y=1$

$1\geq 4(1)-5$

$1\geq -1$ ------> is true

therefore

the point $(1,1)$ is a solution of the inequality

Part 5) Write an inequality to match the graph

we know that

The equation of the line has a negative slope

The y-intercept is the point $(3,0)$

The x-intercept is a positive number

The solution is the shaded area above the dashed line

so

the equation of the line is $y=-5x+3$

The inequality is $y>-5x+3$

3 0

3 years ago

Read 2 more answers

Can somebody help me

wel

Answer:

I believe the second one is d

5 0

3 years ago

On a number line what is the distance between-3/7 and -2/3

Nikitich [7]

Is simply their difference,

$\bf -\cfrac{3}{7}-\left( -\cfrac{2}{3} \right)\implies -\cfrac{3}{7}+\cfrac{2}{3}\impliedby \textit{LCD of 21}\implies \cfrac{-3(3)+2(7)}{21}
\\\\\\
\cfrac{-9+14}{21}\implies \cfrac{5}{21}$

4 0

3 years ago

the parent function of the function g(x)=(x-h)^2+k is f(x)=x^2. the vertex of the function g(x) is located at (9,8). what are th

Lunna [17]

Answer:

<h2>h = 9, k = 8</h2>

Step-by-step explanation:

$\text{The vertex form of an an equation of a quadratic function:}\\\\y=a(x-h)^2+k\\\\(h,\ k)-vertex\\\\\text{We have}\ g(x)=(x-h)^2+k,\ \text{and the vertex in}\ (9,\ 8).\\\\\text{Therefore}\ h=9\ \text{and}\ k=8.\ \text{substitute:}\\\\g(x)=(x-9)^2+8$

8 0

3 years ago

Solve asap I'll mark brainliest

diamong [38]

Answer:

s = 4/21

Step-by-step explanation:

Move all terms not containing s to the right side of the equation, then solve.

5 0

3 years ago

Read 2 more answers