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3 years ago

Explain how you decided where to place the decimal in the quotients for 152÷10, 152÷100,152÷1000

1 answer:

7 0

If the divisor is greater than the dividen then you know that the answer is going to be more than one. But if the divisor is smaller than the dividen then you know that the answer is going to be less than one like a decimal.

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HELP!! PLZ!! I need to get get this right to pass the 6th grade plz help

Harman [31]

23% it shows you what it is right there :)

8 0

3 years ago

Read 2 more answers

What is the answer for this?

ra1l [238]

Answer:

Proved

Step-by-step explanation:

Given

$(2 * 10^2) * (3 * 10^5) = 3.5 * 10^7 + 2.5 * 10^7$

Required

Prove

On the right hand side, we have:

$2 * 10^2 * 3 * 10^5 = 3.5 * 10^7 + 2.5 * 10^7$

Rewrite:

$2 * 3* 10^2 * 10^5 = 3.5 * 10^7 + 2.5 * 10^7$

$6* 10^2 * 10^5 = 3.5 * 10^7 + 2.5 * 10^7$

Apply law of indices

$6* 10^{2+5} = 3.5 * 10^7 + 2.5 * 10^7$

$6* 10^7 = 3.5 * 10^7 + 2.5 * 10^7$

Express 6 as 3.5 + 2.5

$(3.5 + 2.5)* 10^7 = 3.5 * 10^7 + 2.5 * 10^7$

Open bracket

$3.5 *10^7+ 2.5*10^7 = 3.5 * 10^7 + 2.5 * 10^7$

Proved

3 0

3 years ago

Mike has g green marbles and y yellow marbles. Tina keeps losing her marbles. She has half as many yellow and 3 less green marbl

madreJ [45]

there you go ............

5 0

3 years ago

Rectangle 1 has length x and width y. Rectangle 2 is made by multiplying each dimension of Rectangle 1 by a factor of K, where k

STALIN [3.7K]

$\bold{\huge{\underline{\pink{ Solution }}}}$

<h3>Given :- </h3>

Rectangle 1 has length x and width y
Rectangle 2 is made by multiplying each dimensions of rectangle 1 by a factor of k
Where, k > 0

<h3>Answer 1 :-</h3>

Yes, The rectangle 1 and rectangle 2 are similar .

<h3>According to the similarity theorem :-</h3>

If the ratio of length and breath of both the triangles are same then the given triangles are similar.

Let's Understand the above theorem :-

The dimensions of rectangle 1 are x and y

Now, 

Rectangle 2 is made by multiplying each dimensions of rectangle 1 by a factor of k .

Let assume the value of K be 5

Therefore, 

The dimensions of rectangle 2 are

$\sf{ 5x \:and \:5y }$

Now, The ratios of dimensions of both the rectangle :-

$\bold{Rectangle 1 = Rectangle 2}$

$\bold{\dfrac{ x }{y}}{\bold{ = }}{\bold{\dfrac{5x}{5y}}}$

$\bold{\blue{\dfrac{ x }{y}}}{\bold{\blue{ = }}}{\bold{\blue{\dfrac{x}{y}}}}$

From above, 

We can conclude that the ratios of both the rectangles are same

Hence , Both the rectangles are similar

<h3>Answer 2 :- </h3>

Here, 

We have to proof that, the

Perimeter of rectangle 2 = k(perimeter of rectangle 1 )

In the previous questions, we have assume the value of k = 5

<h3>Therefore, </h3>

According to the question,

Perimeter of rectangle 1

$\sf{ = 2( length + Breath) }$

$\bold{\pink{= 2( x + y ) }}$

Thus, The perimeter of rectangle 1

Perimeter of rectangle 2

$\sf{ = 2( length + Breath) }$

$\sf{ = 2(5x + 5y) }$

$\sf{ = 2 × 5( x + y) }$

$\bold{\pink{= 10(x + y) }}$

Thus, The perimeter of rectangle 2

According to the given condition :-

Perimeter of rectangle 2 = k( perimeter of rectangle 1 )

Subsitute the required values,

$\sf{ 2(x + y) = 10(x + y)}$

$\bold{\pink{2x + 2y = 5(2x + 2y) }}$

From Above, 

We can conclude that the, Perimeter of rectangle 2 is 5 times of perimeter of rectangle 1 and we assume the value of k = 5.

Hence, The perimeter of rectangle 2 is k times of rectangle 1

<h3>Answer 3 :</h3>

Here, 

We have to proof that ,

The area of rectangle 2 is k² times of the area of rectangle 1.

That is, 

Area of rectangle 1 = k²( Area of rectangle)

<h3>Therefore, </h3>

According to the question,

Area of rectangle 1

$\sf{ = Length × Breath }$

$\sf{ = x × y }$

$\bold{\red{= xy }}$

Area of rectangle 2

$\sf{ = Length × Breath }$

$\sf{ = 5x × 5y }$

$\bold{\red{ = 25xy }}$

According to the given condition :-

Area of rectangle 1 = k²( Area of rectangle)

$\sf{ xy = 25xy }$

$\bold{\red{xy = (5)²xy }}$

From Above, 

We can conclude that the, Area of rectangle 2 is (5)² times of area of rectangle 1 and we have assumed the value of k = 5

Hence, The Area of rectangle 2 is k times of rectangle 1 .

7 0

2 years ago

Use completing the square to solve for x in the equation (x-12)(x+4)= 9.

iragen [17]

Answer:

x = 4 + sqrt(73) or x = 4 - sqrt(73)

Step-by-step explanation by completing the square:

Solve for x:

(x - 12) (x + 4) = 9

Expand out terms of the left hand side:

x^2 - 8 x - 48 = 9

Add 48 to both sides:

x^2 - 8 x = 57

Add 16 to both sides:

x^2 - 8 x + 16 = 73

Write the left hand side as a square:

(x - 4)^2 = 73

Take the square root of both sides:

x - 4 = sqrt(73) or x - 4 = -sqrt(73)

Add 4 to both sides:

x = 4 + sqrt(73) or x - 4 = -sqrt(73)

Add 4 to both sides:

Answer: x = 4 + sqrt(73) or x = 4 - sqrt(73)

4 0

4 years ago