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2 years ago

The students in a club are selling flowerpots to raise money. Each flowerpot sells for $15.

Mathematics

1 answer:

fiasKO [112]2 years ago

7 0

Answer:

Part A: x·15=$ Part B: Yes

Step-by-step explanation:

Part A: All you have to do is how many pots(x) and multiply it by $15

Part B: 43x15=645 so they do exceed there goal of 500 by 145 dollars

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For the first year of his car insurance policy, Tom pays $77 per month. Tom's payments are reduced to $65 per month in the secon

Tcecarenko [31]

77/65=1.18
round 1.18 to 1.2

6 0

2 years ago

Read 2 more answers

Which statements are true about the graph of y s 3x + 1 and y 2 -x + 2? Check all that apply.

Varvara68 [4.7K]

Let's go through all the options one at a time and look for the correct ones

Option 1: The slope of one boundary line is 2

We have 2 equations of lines, where the coefficients of are 3 and -1 respectively

because the coefficient of x denotes the slope of a line, we know that the lines have the slope 3 and -1, not 2

Hence, this option is Incorrect

Option 2: Both boundary lines are solid

In order for the boundary lines to be solid, the inequality must have an 'equal to', like ≤ (less than or equal to) or ≥ (greater than or equal to)

we can see that that's the case in our case and hence, this option is Correct

Option 3: A solution to the system is (1, 3)

To confirm this, we'll plug these coordinates into the given inequalities and see if it stands correct

y ≤ 3x + 1

3 ≤ 3(1) + 1

3 ≤ 4 which is correct because 3 is less than 4

Second equation:

y ≥ 2 - x

3 ≥ 2 - 1

3 ≥ 1

Which is also true because 3 is greater than 1

Now, we can say that (1 , 3) is a solution to the system because it satisfies both the equations and is Correct

Option 4: Both inequalities are shaded below the boundary lines

For an inequality to be shaded below the boundary line, it must have the ≤ inequality (in case of solid line) and < inequality (in case of dotted line)

because the second inequality listed includes the ≥ inequality, which was not mentioned above, it won't be shaded below

another way to think about it is that any 'greater than' inequality will shade everything above the line and the 'lesser than' inequality will shade below the line

which means that this option is Incorrect

Option 5: The boundary lines intersect

In order for the boundary lines to intersect, they must have have different slopes.

as we mentioned in the explanation of the first option, that the slopes of the lines is 3 and -1, which are different slopes

Therefore, this option is Correct

5 0

1 year ago

Read 2 more answers

"The Highest Common Factor (HCF) of my two numbers is 3 The Lowest Common Multiple (LCM) of my two numbers is 45" (b) Write down

Mila [183]

Answer:

Step-by-step explanation:

15 and 3

6 0

2 years ago

The standard form of the equation of a parabola is x = y2 + 10y + 22. What is the vertex form of the equation?

iVinArrow [24]

From the conics form of the equation, shown above, I look at what's multiplied on the unsquaredpart and see that 4p = 4, so p = 1. Then the focus is one unit above the vertex, at (0, 1), and the directrix is the horizontal line y = –1, one unit below the vertex.

vertex: (0, 0); focus: (0, 1); axis of symmetry: x = 0; directrix: y = –1

Graph y2 + 10y + x + 25 = 0, and state the vertex, focus, axis of symmetry, and directrix.

Since the y is squared in this equation, rather than the x, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking y-values first and then finding the corresponding x-values for x = –y2 – 10y – 25:

To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:

y2 + 10y + 25 = –x
(y + 5)2 = –1(x – 0)

This tells me that 4p = –1, so p = –1/4. Since the parabola opens to the left, then the focus is 1/4 units to the left of the vertex. I can see from the equation above that the vertex is at (h, k) = (0, –5), so then the focus must be at (–1/4, –5). The parabola is sideways, so the axis of symmetry is, too. The directrix, being perpendicular to the axis of symmetry, is then vertical, and is 1/4 units to the right of the vertex. Putting this all together, I get:

vertex: (0, –5); focus: (–1/4, –5); axis of symmetry: y = –5; directrix: x = 1/4

Find the vertex and focus of y2 + 6y + 12x – 15 = 0

The y part is squared, so this is a sideways parabola. I'll get the y stuff by itself on one side of the equation, and then complete the square to convert this to conics form.

y2 + 6y – 15 = –12x
y2 + 6y + 9 – 15 = –12x + 9
(y + 3)2 – 15 = –12x + 9
(y + 3)2 = –12x + 9 + 15 = –12x + 24
(y + 3)2 = –12(x – 2)
(y – (–3))2 = 4(–3)(x – 2)

Then the vertex is at (h, k) = (2, –3) and the value of p is –3. Since y is squared and p is negative, then this is a sideways parabola that opens to the left. This puts the focus 3 units to the left of the vertex.

vertex: (2, –3); focus: (–1, –3)

6 0

1 year ago

Find the dimensions of a deck which will have railings on only three sides. There is 28 m of railing available and the deck must

Roman55 [17]

Answer:

The largest possible area of the deck is 87.11 m² with dimensions;

Width = 9.33 m

Breadth = 9.33 m

Step-by-step explanation:

The area of a given dimension increases as the dimension covers more equidistant dimension from the center, which gives the quadrilateral with largest dimension being that of a square

Given that the railings will be placed on three sides only and the third side will cornered or left open, such that the given length of railing can be shared into three rather than four to increase the area

The length of the given railing = 28 m

The sides of the formed square area by sharing the railing into three while the fourth side is left open are then equal to 28/3 each

The area of a square of side s = s²

The largest possible area of the deck = (28/3)² = 784/9 = 87.11 m² with dimensions;

Width = 28/3 m = 9.33 m

Breadth = 28/3 m = 9.33 m.

5 0

2 years ago