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3 years ago

explain how you solved (4 ten thousands 3 tens)÷10 use a place value chart to support your explanation

Mathematics

1 answer:

OLEGan [10]3 years ago

5 0

This is how it works for every 0 it is divided by you remove a zero or move the decimal point one to the left

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* Use order of operations to solve the following.

Bingel [31]

The answer is 100 if that 5 and 2 is separate

3 0

3 years ago

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Does (0,0) satisfy x+3y>3

PtichkaEL [24]

$\bf x+3y > 3\qquad \qquad (\stackrel{x}{0}~~,~~\stackrel{y}{0})\qquad \qquad 0+3(0) > 3\implies \stackrel{\stackrel{\textit{is this true?}}{\downarrow }}{0>3}\qquad nope!$

4 0

3 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago

Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, -1).

ira [324]

Line:
3x+y=7
y=-3x+7

Slope of Perpendicular Line: 1/3x

For the perpendicular line to contain point (6,-1) the y-intercept would be (0,1), thus the equation of the line would be y=1/3x+1

y=1/3x+1

5 0

3 years ago

3 + 4x - 2x, when x = 0.5

IgorLugansk [536]

Replace x with 0.5
3+4(0.5)-2(0.5)
Now multiply
3+2-1
5-1
Answer: 4

8 0

3 years ago

Read 2 more answers