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3 years ago

Ned had a lump of play doh that was 31/2 inches long. If he stretched it out 2 times it current length how long would it be

Mathematics

2 answers:

viktelen [127]3 years ago

7 0

Answer:

It would be 7 inches long.

Step-by-step explanation:

If he stretched it out 2 times the current length that means that you need to multiply. 3 1/2 can be put into the calculator as 3.5. 3.5 multiplied by 2 equals 7.

choli [55]3 years ago

4 0

Answer: 7

Step-by-step explanation:

3.5 times 2 is 7

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John bought an antique model car priced at $72 she also had to pay 5% sales tax what was the total amount she paid

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Multiply $72 into 5% and that should give you an answer

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3 years ago

Solve the system of equations. \begin{aligned} & -10x+3y = 5 \\\\ & x=y-4 \end{aligned} −10x+3y=5 x=y−4

Ksenya-84 [330]

Answer:

x = 1

y = 5

Step-by-step explanation:

You can use either substitution or elimination for this problem. I will use substitution.

Step 1: Replace x in the 1st equation

-10(y-4) + 3y = 5

-10y + 40 + 3y = 5

-7y = -35

y = 5

Step 2: Plug y into an original equation to find x

x = 5 - 4

x = 1

And you have your final answers!

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3 years ago

What is 2/3 cup chopped nuts multiplied by 6 ?

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12/3 cup chopped nuts
which simplified is 4 cups

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3 years ago

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tatuchka [14]

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Step-by-step explanation:

$1000 x .12 =120/2=60

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3 years ago

Which equation is y = –6x^2 + 3x + 2 rewritten in vertex form? y = negative 6 (x minus 1) squared + 8 y = negative 6 (x + one-fo

mart [117]

Answer:

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

Step-by-step explanation:

Given:

$y = -6x^2 + 3x + 2$

Required

Rewrite in vertex form

The vertex form of an equation is in form of: $y = a(x - h)^2+ k$

Solving: $y = -6x^2 + 3x + 2$

Subtract 2 from both sides

$y - 2 = -6x^2 + 3x + 2 - 2$

$y - 2 = -6x^2 + 3x$

Factorize expression on the right hand side by dividing through by the coefficient of x²

$y - 2 = -6(x^2 + \frac{3x}{-6})$

$y - 2 = -6(x^2 - \frac{3x}{6})$

$y - 2 = -6(x^2 - \frac{x}{2})$

Get a perfect square of coefficient of x; then add to both sides

------------------------------------------------------------------------------------

Rough work

The coefficient of x is $\frac{-1}{2}$

It's square is $(\frac{-1}{2})^2 = \frac{1}{4}$

Adding inside the bracket of $-6(x^2 - \frac{x}{2})$ to give: $-6(x^2 - \frac{x}{2} + \frac{1}{4})$

To balance the equation, the same expression must be added to the other side of the equation;

Equivalent expression is: $-6(\frac{1}{4})$

------------------------------------------------------------------------------------

The expression becomes

$y - 2 -6(\frac{1}{4})= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{6}{4}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{3}{2}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

Factorize the expression on the right hand side

$y - 2 -\frac{3}{2}= -6(x - \frac{1}{2})^2$

$y - (2 +\frac{3}{2})= -6(x - \frac{1}{2})^2$

$y - (\frac{4+3}{2})= -6(x - \frac{1}{2})^2$

$y - (\frac{7}{2})= -6(x - \frac{1}{2})^2$

$y +\frac{7}{2} = -6(x - \frac{1}{2})^2$

Make y the subject of formula

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

Solved

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3 years ago

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