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3 years ago

7

Which number variable expression represents the number of inches in y yards?

2 answers:

wlad13 [49]3 years ago

4 0

Answer=36*y

36 inches in 1 yard
36in=1yd

36*y

DerKrebs [107]3 years ago

4 0

To express your answer.... the answer is 36*y.

36*y (there are 36 inches in 1 yard)

36in = 1 yd

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Brian is finishing a meal at his favorite restaurant. The check shows that the cost of the food is $42. He adds a 15% tip to the

Rus_ich [418]

Answer:

48.30

Step-by-step explanation:

First determine the tip

42 *15%

42 *.15 =6.30

Add this to the amount of the tip

42 + 6.30 =48.30

The total amount is 48.30

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3 years ago

A space shuttle is made

zaharov [31]

Answer:

The actual space shuttle is 66 feet tall

Step-by-step explanation:

scale of 1 inch : 12 feet represent the scale of a space shuttle.

It means it has a representation of 1 inch on paper then actual = 12 feet.

But then if the paper(model) changes to 5.5 inches. The actual representation on land will be

Let me write out the equation

1 inch = 12 feet

5.5 inch =( 5.5 inch *12 feet)/1 inch

5.5 Inch= 5.5 * 12 feet

5.5 inch = 66 feet

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3 years ago

Find the value of the variable y, where the sum of the fractions 6/(y+1) and y/(y-2) is equal to their product.

Blizzard [7]

Answer:

The answer is

$y = 3$

$y = - 4$

Step-by-step explanation:

We must find a solution where

$\frac{6}{y + 1} + \frac{y}{y - 2} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Consider the Left Side:

First, to add fraction multiply each fraction on the left by it corresponding denomiator and we should get

$\frac{6}{y + 1} \times \frac{y - 2}{y - 2} + \frac{y}{y - 2} \times \frac{y + 1}{y + 1}$

Which equals

$\frac{6y - 12}{(y -2) (y + 1)} + \frac{ {y}^{2} + y }{(y - 2)(y + 1)}$

Add the fractions

$\frac{y {}^{2} + 7y - 12 }{(y - 2)(y + 1)} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Simplify the right side by multiplying the fraction

$\frac{6y}{(y + 1)(y + 2)}$

Set both fractions equal to each other

$\frac{6y}{(y + 1)(y - 2)} = \frac{ {y}^{2} + 7y - 12}{(y + 1)(y - 2)}$

Since the denomiator are equal, we must set the numerator equal to each other

$6y = {y}^{2} + 7y - 12$

$= {y}^{2} + y - 12$

$(y + 4)(y - 3)$

$y = - 4$

$y = 3$

6 0

3 years ago

Read 2 more answers

This is my question

vovangra [49]

Answer:

week 4 had the greatest weight change

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3 years ago

let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep

Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

$x_1,x_2$ and $x_3$ are linearly independent.

By definition of linear independent there exits three scalar $a_1,a_2$ and $a_3$ such that

$a_1x_1+a_2x_2+a_3x_3=0$

Where $a_1=a_2=a_3=0$

$y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1$

We have to prove that $y_1,y_2$ and $y_3$ are linearly independent.

Let $b_1,b_2$ and $b_3$ such that

$b_1y_1+b_2y_2+b_3y_3=0$

$b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0$

$b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0$

$(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0$

$b_1+b_3=0$

$b_1=-b_3$ ...(1)

$b_1+b_2=0$

$b_1=-b_2$ ..(2)

$b_2+b_3=0$

$b_2=-b_3$ ..(3)

Because $x_1,x_2$ and $x_3$ are linearly independent.

From equation (1) and (3)

$b_1=b_2$ ...(4)

Adding equation (2) and (4)

$2b_1==0$

$b_1=0$

From equation (1) and (2)

$b_3=0,b_2=0,b_3=0$

Hence, $y_1,y_2$ and $y_3$ area linearly independent.

5 0

3 years ago