Divide two and one 50 x 0.1

Kiera used place value and the distributive property to find the product of 12 and 597. Her work is shown below. 12(597) = 12(50

olya-2409 [2.1K]

Answer:

(90)(7) ⇒ the error was multiplying 90 and 7.

Step-by-step explanation:

12(597) = 12(500 + 90 + 7)

= 12(500) + (90)(7)

= 6000 + 630

= 6,630

The part in bold shows where the error is. It was not correct to multiply 90 and 7.

The process was to proceed as follows:

12(597) = 12(500 + 90 + 7)

= 12(500) + 12(90) + 12(7)

= 6000 + 1080 + 84

= 7,164

8 0

3 years ago

What is the equation of this line? y=−3/2x y=2/3x y=3/2x y=−2/3x Graph of a line passing through the origin and the point begin

iogann1982 [59]

Correct answer: y=(2/3)x

The line passes through the origin (0,0) and the pair (3,2)

the equation y=(2/3)x satisfies both points:

for (0,0): 0=(2/3)*0 = 0

for (3,2): 2 = (2/3)*3 = 2

and therefore is the correct linear function of the line described.

4 0

3 years ago

Read 2 more answers

Answer please!!!!!!!!!!!!!!!!!!

castortr0y [4]

The rule is "whatever the input is, multiply it by 5, then subtract 2"

For example, if the input is 4 then 4*5 = 20 and 20-2 = 18
So in short, the input 4 leads to the output 18 as shown in the top row of the diagram.

The answer is the box on the bottom row, right hand side where it says "x 5" and "-2" in the two bubbles.

7 0

4 years ago

Marie is reading a 271-page book. She has already read 119 pages. She uses the equation 119+8h=271 to find out how long it will

VikaD [51]

Answer:

12.3

Step-by-step explanation:

271-119= 152

get square root of 152

12.3x12.3=152

5 0

3 years ago

Read 2 more answers

Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions:

grigory [225]

Answer:

Part A: The graph of the system is plotted at y-intercepts of 3 and -4. The line of the first equation is dashed and shaded below. The line of the second equation is dashed and shaded above. The solution area is between the second and third quadrants.

Part B: No, the point (-4, 6) is not in the solution are because it only fits into the first equations solution area. [6=2(-4)+3 solved is 6=-5, this is not a true statement.]

Step-by-step explanation:

Part A: Each equation {y>2x+3, y<-3/2x-4} is already in slope intercept form and can be graphed. In the first equation we know that there is a y-intercept of 3 and a slope of 2. This is shown as as a upward moving line with plot points such as (-2,-1), (-1,1), (0,4), (1,6), and so on. In the second equation we know that there is a y-intercept of -4 and a slope of -3/2. This is shown as a downward moving line with plot points such as (0,-4), (-2,-1), (-4, 2), and so on. (See graph below)

Part B: The plot point (-4, 6) is not included in the solution area of the systems. This can be seen on the graph. And, proven mathematically by plugging in the point into the equations:

y=2x+3 plug in the points (6)=2(-4)+3
solve (6)=2(-4)+3 ... 2(-4)=-8 ... -8+3=-5 ... 6 = -5 is not a true statement

So, already we know that this plot point is not an answer to the system. Let's continue anyway.

y=-3/2x-4 plug in the points (6)=-3/2(-4)-4
solve (6)=-3/2(-4)-4 ... -3/2(-4)= 6 ... 6-4 = 2 ... 6=2 is not a true statement.

7 0

3 years ago