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3 years ago

What is the greatest 3 digit number you can subtract from 426 so that you would need to regroup?

Mathematics

2 answers:

AURORKA [14]3 years ago

7 0

Answer:

419

Step-by-step explanation:

Regrouping in subtraction: If we borrow from one group (tens group) and give to another (Once group) so that the operation can be completed, then this process is called regroup.

For example: We need to find the value of 52-39

$52=50+2$

$39=30+9$

We can not subtract 9 from 2. So we will borrow 10 from 50 and give to 2.

$52=40+12$

Now,

$52-39=(40+12)-(30+9)$

$52-39=(40-30)+(12-9)$

$52-39=(10)+(3)$

$52-39=13$

We need to find the greatest 3 digit number that can subtract from 426 so that we need to regroup.

To use regroup, the once digit must be more that 6. So, maximum value for once place is 9.

Therefore, the greatest 3 digit number that can subtract from 426 by using regroup is 419.

kolbaska11 [484]3 years ago

6 0

Do you want me to explain it?

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Two boats depart from a port located at (–8, 1) in a coordinate system measured in kilometers and travel in a positive x-directi

miss Akunina [59]

Answer:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

Step-by-step explanation:

1st boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=1\\ \\b=-2a$

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$y=ax^2 -2ax+c$

The y-coordinate of the vertex:

$y_v=a\cdot 1^2-2a\cdot 1+c\Rightarrow a-2a+c=10\\ \\c-a=10$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-2a\cdot (-8)+c\\ \\80a+c=1$

Solve:

$c=10+a\\ \\80a+10+a=1\\ \\81a=-9\\ \\a=-\dfrac{1}{9}\\ \\b=-2a=\dfrac{2}{9}\\ \\c=10-\dfrac{1}{9}=\dfrac{89}{9}$

Parabola equation:

$y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}$

2nd boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=0\\ \\b=0$

Equation:

$y=ax^2+c$

The y-coordinate of the vertex:

$y_v=a\cdot 0^2+c\Rightarrow c=-7$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-7\\ \\64a-7=1$

Solve:

$a=-\dfrac{1}{8}\\ \\b=0\\ \\c=-7$

Parabola equation:

$y=\dfrac{1}{8}x^2 -7$

System of two equations:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

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Answer:

12) A = 27

13) B = 36

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Step-by-step explanation:

We need to complete the ratio tables

We are given:

Footballs 15 5 20 C D

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The table shows the ratio of footballs to baseballs.

We are given the ratio $\frac{Football}{Baseball}=\frac{5}{9}$

All other ratios will also be simplified to get $\frac{5}{9}$

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We know that $\frac{5\times3}{9\times3}= \frac{15}{27}$

So, A = 27

13) Calculate B

Footballs = 20

Baseballs = B

Common ratio is 5/9, we need to find a number that multiplies by 5 to get 20, same number will be multiplied by denominator to get B.

We know that $\frac{5\times4}{9\times4}= \frac{20}{36}$

So, B = 36

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Common ratio is 5/9, we need to find a number that multiplies by 9 to get 45, same number will be multiplied by numerator to get C.

We know that $\frac{5\times5}{9\times5}= \frac{25}{45}$

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