1) Which two lines are parallel?

I. 5y = -3x-5
II. 5y = -1 - 3x
III. 3y - 2x = -1


a. I. and II.
b. I. and III.
c. II. and III.
d. No two of the lines are parallel.



2) Which do lines are parallel?

I. 3y = -2x - 3
II. 3y = 1 - 2x
III. 4y - 3x = -1



a. I. and II.
c. II. and III.
b. I. and III.
d. No two of the lines are parallel.

The area of the triangle is half the product of its base and height. The area of the triangle is 137.25 in².

The area of the triangle is half the product of its base and height. It is given by the formula,

[tex]\rm \text{Area of triangle} = \dfrac{1}{2} \times base \times height[/tex]

As it is given that the base of the triangle is 15¹/₄ inches, while the height of the triangle is 18 inches. Therefore, the area of the triangle is,

[tex]\rm \text{Area of triangle} = \dfrac{1}{2} \times base \times height[/tex]

                         [tex]= \dfrac{1}{2} \times 15(1/4)\times 18\\\\\\ = \dfrac{1}{2} \times \dfrac{(15\times 4)+1}{4} \times 18\\\\\\= \dfrac{1}{2} \times \dfrac{61}{4} \times 18\\\\\\=137.25\rm\ in^2[/tex]

Hence, the area of the triangle is 137.25 in².

Learn more about the Area of the triangle:

https://brainly.com/question/22457738

To find the value of X in a rectangle with width X and length 2X, we can use the area formula. By solving the equation, we find X = 9, making the width of the rectangle 9 units.

To find the value of X, we need to use the given information about the length and width of the rectangle. Let's assume the width of the rectangle is X. Since the length is double the width, it is 2X. The area of the rectangle is given as 162, so we can set up the equation:

Width × Length = Area

X × 2X = 162

Simplifying the equation, we get:

2X^2 = 162

Dividing both sides by 2, we have:

X^2 = 81

Taking the square root of both sides, we find:

X = ±9

Since the width cannot be negative, the value of X is 9. Therefore, the width of the rectangle is 9 units.

Learn more about Finding the width of a rectangle here:

https://brainly.com/question/34224051

#SPJ2

The width of the rectangle is 9 units.

To find the width of the rectangle, we can set up the equation for the area. The width of the rectangle is X, and the length is twice the width, meaning the length is 2X. The formula for the area is length times width, so we have X * 2X = 162. Simplifying this equation gives us 2X^2 = 162. Dividing both sides by 2 gives us X^2 = 81. Taking the square root of both sides gives us X = 9. Therefore, the width of the rectangle is 9 units.

https://brainly.com/question/26613529

#SPJ5

Consider the difference of polynomials [tex](7x^8 + 8x^6 - 2x + 9) - (4x^8 + 3x^7 + 3).[/tex]

The first thing you have to do is take that "minus" sign through the parentheses containing the second polynomial (just change sign + to - and sign - to +):

[tex](7x^8 + 8x^6 - 2x + 9) - (4x^8 + 3x^7 + 3)=7x^8+8x^6-2x+9-4x^8-3x^7-3.[/tex]

Then combine terms with the same degree:

[tex]7x^8+8x^6-2x+9-4x^8-3x^7-3=(7x^8-4x^8)-3x^7+8x^6-2x+(9-3).[/tex]

Note that

[tex]7x^8-4x^8=(7-4)x^8=3x^8,\\ \\9-3=6.[/tex]

Thus,

[tex](7x^8 + 8x^6 - 2x + 9) - (4x^8 + 3x^7 + 3)=3x^8-3x^7+8x^6-2x+6.[/tex]

Answer:

Brandon is 13 and Ryan is 6 and 1/2

Step-by-step explanation

Answer:

  $55

Step-by-step explanation:

Put 8 where n is in the equation and do the arithmetic.

  cost = $6.50·n + 3

  = $6.50·8 + 3 = $52 + 3

  cost = $55

The number that is ten more than 43 is 53

Represent the number with x.

A number x that is 10 more than 43 is then calculated using the following equation

x - 43 = 10

Add 43 to both sides of the equation

x - 43 + 43 = 10 + 43

Evaluate the sums

x = 53

Hence, the number that is ten more than 43 is 53

See attachment for the place value blocks

Read more about place values at:

https://brainly.com/question/12386995

To find the number ten more than 43, use place-value blocks by adding a tens block to the existing four tens and three ones blocks to get 53. On a hundred chart, locate the number 43 and move one row down to arrive at 53, which is ten more than 43.

The question asks how to use place-value blocks and a hundred chart to show the number that is ten more than 43. To illustrate this using place-value blocks, you would start with 4 tens blocks and 3 ones blocks to represent the number 43. Then, you would add one more tens block to represent the addition of ten. After adding this block, you would have a total of 5 tens blocks and 3 ones blocks, which represents the number 53.

On a hundred chart, first find the number 43. Then, since each row on a hundred chart represents ten numbers, you would move one row down to find the number that is ten more than 43. This would take you to the number 53, which is directly below 43 on the chart.

Therefore, the number that is ten more than 43 is 53.

Gloria went to the mall and spent 50 minutes shopping. Then she had lunch for 30 minutes. If Gloria arrived at the mall at 11:00 A.M., at what time did she finish lunch?

Answer:

The correct option is 4.

Step-by-step explanation:

The given expression is

[tex]25a+5b-13[/tex]

In option (1), the given expression is

[tex]20a+5a+b-13[/tex]

Combine like terms.

[tex]25a+b-13[/tex]

This expression is not equivalent to the given expression.

In option (2), the given expression is

[tex]20(5a+5b)-13[/tex]

Using distributive property.

[tex]100a+100b-13[/tex]

This expression is not equivalent to the given expression.

In option (3), the given expression is

[tex]30ab-13[/tex]

This expression is not equivalent to the given expression.

In option (4), the given expression is

[tex]5(5a+b)-13[/tex]

Using distributive property.

[tex]25a+5b-13[/tex]

This expression is equivalent to the given expression.

Therefore correct option is 4.