NEEEDDDDD HELPPPPP ASAPPPPP PLZZZZ!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Hector is training for a race that is 10 kilometers long. On Saturday he ran 2.5 km. He plans to add 0.5 km to his distance every Saturday. Hector's Calculations: 2.5 + 0.5 = 3.0 6.5 + 0.5 = 7.0 3.0 + 0.5 = 3.5 7.0 + 0.5 = 7.5 3.5 + 0.5 = 4.0 7.5 + 0.5 = 8.0 4.0 + 0.5 = 4.5 8.0 + 0.5 = 8.5 4.5 + 0.5 = 5.0 8.5 + 0.5 = 9.0 5.0 + 0.5 = 6.0 9.0 + 0.5 = 9.5 6.0 + 0.5 = 6.5 9.5 + 0.5 = 10.0 The equation equivalent to Hector’s calculations is 2.5 + 0.5x = 10. Use either of these methods to determine how many weeks it will take Hector to reach his goal of a 10 kilometer run.

When lithium and phosphorus combine to form lithium phosphide, three lithium cations (Li+) combine with one phosphide anion (P₃-) to balance the charges and form a neutral compound, Li₃P.

an ionic compound is a chemical compound composed of ions held together by electrostatic forces termed ionic bonding.

The correct description of the ionic compound lithium phosphide, with the chemical formula Li3P, is "Lithium ions have 1 positive charge, and phosphorus ions have 3 negative charges."

This is because lithium is a Group 1A element, which means it has one valence electron that it tends to lose in chemical reactions to form a cation with a charge of +1.

Phosphorus, on the other hand, is a Group 5A element, which means it has five valence electrons and tends to gain three electrons to form an anion with a charge of -3.

Thus, When lithium and phosphorus combine to form lithium phosphide, three lithium cations (Li+) combine with one phosphide anion (P₃-) to balance the charges and form a neutral compound, Li₃P.

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The length of the segment formed by the two given endpoints with coordinates coordinates (5, 2) and (–6, 2) is;

d = 11

Formula for distance between two coordinates is given as;

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

We are given the coordinates (5, 2) and (–6, 2).

This means that;

x₁ = 5 and x₂ = -6

y₁ = 2 and y₂ = 2

Thus;

d =  √[(-6 - 5)² + (2 - 2)²]

d = √(121 + 0)

d = √121

d = 11

      In conclusion, the length of the segment formed by the two given endpoints is; d = 11

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In a rhombus, the diagonals are perpendicular bisectors of each other, meaning they divide each other into two equal parts at a right angle. Let's denote the measure of ∠BAD as x degrees. The measure of ∠BAD is [tex]\( 26^\circ \)[/tex].

Since the angle bisector of ∠ACD makes a 64° angle with the diagonal BD, and BD bisects ∠ACD, then ∠ACB = 64° as well. Additionally, since ABCD is a rhombus, opposite angles are equal. Therefore, ∠ABD = ∠ACB = 64°.

Now, considering triangle ABD, we know that the sum of angles in a triangle is 180°. Thus:

[tex]\[ \angle BAD + \angle ABD + \angle ADB = 180^\circ \][/tex]

Substituting the known values:

[tex]\[ x + 64^\circ + 90^\circ = 180^\circ \][/tex]

[tex]\[ x + 154^\circ = 180^\circ \][/tex]

[tex]\[ x = 180^\circ - 154^\circ \][/tex]

[tex]\[ x = 26^\circ \][/tex]

So, the measure of ∠BAD is [tex]\( 26^\circ \)[/tex].

To summarize:

1. The angle bisector of ∠ACD in the rhombus makes a 64° angle with diagonal BD.

2. By the properties of a rhombus, ∠ACB = 64° as well.

3. Since opposite angles in a rhombus are equal, ∠ABD = ∠ACB = 64°.

4. In triangle ABD, the sum of angles is 180°.

5. Solving for ∠BAD, we find [tex]\( 26^\circ \).[/tex]

Answer:I = 550 × 0.03 × 1

Step-by-step explanation: from plato

Final answer:

To share the ribbon equally among the 3 friends, each person will get 1/12 yard of ribbon.

Explanation:

To determine how much ribbon each person will get, we need to divide the total length of the ribbon equally among all friends.

1/4 yard of ribbon divided by 3 friends gives each person 1/12 yard of ribbon.

Therefore, each person will get 1/12 yard of ribbon for breast cancer awareness month.

The given graph is for a piecewise function consisting of two intervals:

(1) x ≤ 1 ⇒⇒⇒ the graph in this interval like a cubic function.

(2) x > 1 ⇒⇒⇒ the graph in this interval like a quadratic function

By comparing to the given options

The answer is option c.

Answer:

Option C is correct.

Explanation:

A piece-wise function is a function which is defined by multiple sub-functions, each sub-function applying to a certain interval of the main function's domain.

From the graph we can see that the function is made up of 2 pieces.

Solid dot means- Including

Open-dot means - not- including.

As we can see that the boundary lines are at x=1.

We have to look each line separately to determine each equation:

when [tex]x>1[/tex], it gives the function [tex]y=x^2+4[/tex] and

when [tex]x\leq 1[/tex], it gives the function [tex]y =x^3-2[/tex].

So, the piece-wise function is:

[tex]y=\left \{ {{x^3-2}, x\leq1 \atop {x^2+4}, x>1} \right.[/tex]

The value of Surface area of right cone with radius of 8 and 14 height is,

⇒ SA = 606.31

To multiply means to add a number to itself a particular number of times. Multiplication can be viewed as a process of repeated addition.

Given that;

Radius of cone = 8

And, Height = 14

We know that;

Surface area of right cone is,

SA = πr (r + √h² + r²)

SA = 3.14 × 8 (8 + √14² + 8²)

SA = 606.31

Thus, The value of Surface area of right cone with radius of 8 and 14 height is,

⇒ SA = 606.31

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To determine the number of small and large IC chips sold, we set up a system of linear equations with the given total number of chips (2200) and total earnings ($5050). Solving the system, we found that 1500 small IC chips and 700 large IC chips were sold.

To solve the problem, we can set up a system of linear equations to represent the sales for small and large IC chips. We have two unknowns: the number of small IC chips (let's call it x) and the number of large IC chips (let's call it y). We have two pieces of information to create our equations:

The total number of chips sold is 2200, so x + y = 2200.

The total amount collected is $5050, so 1.50x + 4.00y = 5050.

Solving this system of equations, we multiply the first equation by 1.50 to align it with the second equation:

1.50x + 1.50y = 3300 (Multiply the first equation)

1.50x + 4.00y = 5050 (Second equation)

Subtracting the first new equation from the second gives us:

2.50y = 1750

Dividing both sides by 2.50 to find y gives us y = 700. That means 700 large IC chips were sold. Now, we can substitute y in the first equation to find x:

x + 700 = 2200

x = 2200 - 700

x = 1500

Hence, 1500 small IC chips and 700 large IC chips were sold.

Answer:

The measure of TU is 15.

Step-by-step explanation:

Karla's book is 280 pages long because 85% of the book represents 238 pages, which means the entire book is 100% of that amount calculated by dividing 238 by 0.85. The additional questions can't be addressed accurately without more context or information.

If Karla has read 85% of her book and that represents 238 pages, we can calculate the total length of the book. To find the full length of the book, we need to determine 100% of the book when 85% equals 238 pages. We can set up a proportion or use the formula Part = Percent \times Whole to solve for the whole, which is the total length of the book.

Let's use the formula to find the total length of the book:

Therefore, the total length of Karla's book is 280 pages.

Regarding the additional questions:

Without any calculations, it's not likely that the professor will need to read a total of more than 1,050 pages if we are only referencing the pages that Karla has read. However, if we are considering a separate piece of text or assignment for the professor, a calculation would be necessary to determine the likelihood.

The idea of dividing the total pages by the number of available days is a good strategy for managing reading assignments. This technique helps to break down the task into manageable daily portions and allows better time management.

As for the probability that the professor will need to read more than 1,050 pages, we cannot calculate this without additional information about the total number of assignments and the length of each assignment. Moreover, it is unlikely that the average length of the papers will be less than 12 pages if we consider typical lengths for academic papers, which often exceed this length.