Is 5 a prime composite or neither

Answer: C - Y = -16(x- 2)^2 + 68

Step-by-step explanation:

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Final answer:

The correct quadratic equation that models the ball's vertical motion is y = –16(x – 2)² + 68, since it reflects the ball reaching its maximum height at 2 seconds and the vertex of the parabola being the highest point of the ball's path.

Explanation:

The question involves finding an equation that models the height of a tennis ball thrown straight up into the air after a certain number of seconds. Given that the ball reaches its maximum height at 2 seconds, we can determine which equation best describes the ball's vertical motion. The correct equation will show the ball peaking at 2 seconds and then descending symmetrically in a parabolic path.

The equation that correctly models the motion of the ball in this context is y = –16(x – 2)² + 68. This quadratic equation is in the vertex form of a parabola, where the maximum height is represented by the vertex point (2, 68), and the vertex is the highest point of the parabola since the coefficient of the squared term is negative.

When x is 2 seconds, the equation simplifies to y = –16(0)²+ 68, which results in y = 68, demonstrating that the maximum height of 68 is indeed reached at 2 seconds after the ball is thrown, thus supporting that this is the correct quadratic equation for the scenario.

Therefore, the correct option is C. 15/18 = 25/30. The fractions 15/18 and 25/30 were simplified to 5/6, proving they are equal.

Comparison of Fractions 15/18 and 25/30

To compare the fractions 15/18 and 25/30, we can first simplify them to make the comparison easier. We simplify each fraction by dividing the numerator and denominator by their greatest common divisor (GCD).

1. Simplify 15/18:

GCD of 15 and 18 is 3.

→ 15 ÷ 3 = 5

→ 18 ÷ 3 = 6

So, 15/18 = 5/6.

2. Simplify 25/30:

GCD of 25 and 30 is 5.

→ 25 ÷ 5 = 5

→ 30 ÷ 5 = 6

So, 25/30 = 5/6.

Therefore, 15/18 = 25/30. The correct comparison is:

C. 15/18 = 25/30

Answer:

B.False.

Step-by-step explanation:

A secant  is a line or segment  which passes through two points or more points and cut the curve at atleast two points.

We are talking about of secant of circle

Secant: A secant is a line that passe through two points of circle and it  intersect the circle at two points.

Therefore,given definition of secant is

A secant is a line or segment  that passes through a point on the circle and the cente rof the circle.

By definition of secant , we can say the given define in question is false.

Because that  segment which passes through a point on circle and center of the circle is radius not secant.

The greatest common factor (GCF) of the expressions 2^5 · 5· 11 and 2^3· 5^2 · 7 is 2^3 · 5, or 40. This is determined by comparing the exponents of the common base terms in both expressions and selecting the lowest exponents.

The student's question is about finding the greatest common factor (GCF) of two different mathematical expressions, namely: 2^5 · 5· 11 and 2^3· 5^2 · 7. In this context, 'GCF' refers to the greatest number that is a factor of two or more numbers. For these expressions, we will look at the shared factors of the two expressions.

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a. the total change in the size of the tumor during the first 6 months is 64 - 1 = 63 mm³.

b. the average rate of change in the size of the tumor during the first 6 months is 10.5 mm³ per month.

c. the estimated rate at which the tumor is growing at t = 6 is approximately 44.352 mm³ per month.

a) Total change in the size of the tumor during the first 6 months:

To find the total change in the size of the tumor, we need to subtract the initial size from the size after 6 months. Since [tex]\( S = 2^t \)[/tex], we can calculate the size at t = 6 by plugging in t = 6 into the equation. Then, we can subtract the initial size from this value.

[tex]\[ S(6) = 2^6 = 64 \][/tex]

The initial size is given by [tex]\( S(0) = 2^0 = 1 \).[/tex]

So, the total change in the size of the tumor during the first 6 months is 64 - 1 = 63 mm³.

b) Average rate of change in the size of the tumor during the first 6 months:

The average rate of change is given by the total change divided by the number of months. We already found the total change (63 mm³), and the number of months is 6.

[tex]\[ \text{Average rate of change} = \frac{\text{Total change}}{\text{Number of months}} = \frac{63}{6} = 10.5 \][/tex]

So, the average rate of change in the size of the tumor during the first 6 months is 10.5 mm³ per month.

c) Estimate the rate at which the tumor is growing at t = 6:

To estimate the rate at which the tumor is growing at t = 6, we can use calculus to find the derivative of the function [tex]\( S(t) = 2^t \)[/tex] with respect to t. The derivative represents the rate of change of S with respect to t at any given time.

[tex]\[ \frac{dS}{dt} = \frac{d}{dt} (2^t) = (\ln 2) \cdot (2^t) \][/tex]

Now, plug in t = 6 to find the rate of change at that specific time.

[tex]\[ \frac{dS}{dt}\bigg|_{t=6} = (\ln 2) \cdot (2^6) \][/tex]

[tex]\[ \frac{dS}{dt}\bigg|_{t=6} = (\ln 2) \cdot 64 \][/tex]

Since [tex]\( \ln 2 \approx 0.693 \)[/tex], we have:

[tex]\[ \frac{dS}{dt}\bigg|_{t=6} \approx 0.693 \cdot 64 \approx 44.352 \][/tex]

So, the estimated rate at which the tumor is growing at t = 6 is approximately 44.352 mm³ per month.

40% = 80,000

20% = 40,000

x5 = 100% = 200,000

Keywords:

Functions, subtraction, operation of functions

For this case, we are given two functions of the form [tex]y_ {1} = f (x)[/tex], where [tex]f (x) = 4x + 1[/tex] and[tex]y_ {2} = g (x)[/tex], where [tex]g (x) = x ^ 2-5[/tex]. We must find the subtraction (f-g) (x). So, by definition of operation of functions we have to:

[tex](f-g) (x) = f (x) -g (x)[/tex]

So, we have:

[tex](f-g) (x) = 4x + 1- (x ^ 2-5)[/tex]

Knowing that: [tex]- * + = -\ and\ - * - = +[/tex]

[tex](f-g) (x) = 4x + 1-x ^ 2 + 5\\(f-g) (x) = - x ^ 2 + 4x + 6[/tex]

Answer:

[tex](f-g) (x)[/tex]is given by[tex]-x ^ 2 + 4x + 6[/tex]

Answer: 0.3 is the mode

Step-by-step explanation: The mode is what number shows up the most and 0.3 shows up 3 times.

0 shows up twice

The slope doesn't change after dilation, it would still be 2.

Differential equation: In solving equations involving x, y, and z, taking steps with various values of h optimizes the process.

Step size h: When h=0.5 over the interval 0 to 2, solutions can be calculated by incrementing x from 1 to 1+h.

Optimizing procedures: By choosing appropriate values for h, the iterative process can be streamlined for better efficiency.

The construction method for creating a perpendicular line through a point not on a line is indeed useful for constructing a perpendicular bisector, based on the principle that all right angles are equal to one another. Therefore given statement is true.

The construction method used to construct a perpendicular through a point not on a line can indeed be used to construct a perpendicular bisector. This is true because both constructions rely on the same geometric principles. E4, which states that all right angles are equal to one another, supports this fact. In constructing a perpendicular line through a point not on the line, one essentially finds the point on the existing line that is closest to the external point and constructs a right angle there. This same method can be adapted for a segment to create a perpendicular bisector. By selecting the midpoint of the segment and then constructing a right angle using the segment as one side, this similarly results in a perpendicular bisector.

Final answer:

The construction method used to construct a perpendicular through a point not on a line being also applicable for constructing a perpendicular bisector is True. Both involve similar steps of drawing arcs with a compass that intersect and then drawing a line through those intersections, leveraging the properties of equidistant points and perpendicular bisectors.

Explanation:

The statement that the construction method used to construct a perpendicular through a point not on a line could also be used to construct a perpendicular bisector is True. The basic geometrical constructions used for both are quite similar. To construct a perpendicular through a point not on a line, one would typically use the following process:

Place the compass at the given point and draw an arc that crosses the line twice.

Without changing the compass width, move the compass to each intersection point and draw two arcs that intersect above and below the line.

Draw a line connecting the intersection points of these arcs. This is the perpendicular line through the point.

To construct a perpendicular bisector, one would:

Place the compass at one end of the line segment and draw an arc above and below the line.

Without changing the compass width, repeat from the other end of the line segment so that the arcs cross.

Connect the intersection points of the arcs to form the perpendicular bisector.

In both cases, we utilize the property that the perpendicular bisector of a line segment is the set of all points equidistant from the segment's endpoints, and similarly, through a point not on a line, we can construct two such points that are equidistant from it on the line, therefore satisfying Corollary regarding points equidistant from the extremities of a straight line determining a perpendicular to the line at its middle point.

Answer:

9 is the right answer on edge

Step-by-step explanation:

Break apart the factor 8 into 5 and 3, and then use the distributive property to calculate 7x5 and 7x3. Add the two products together to find that the product of 7 and 8 is 56.

To find the product of 7 and 8, you can break apart the array into more manageable pieces. One way to do this is by using the distributive property, which in this case lets you split one of the factors and multiply each part separately before adding the results. For example, you can break the number 8 into 5 and 3 because 5 + 3 equals 8. Then you multiply 7 by each part:

After calculating these two products, you add them together to find the final product:

35 + 21 = 56

Therefore, the product of 7 and 8 is 56. This approach can make multiplication easier, especially when dealing with larger numbers or when trying to calculate in your head.

To find the standardized version of x in a given city's annual precipitation data, use Z = (x - mean) / standard deviation.

The standardized version of x (annual precipitation in a city) can be calculated using the formula: Z = (x - mean) / standard deviation. Substituting the given values, Z = (x - 15.3) / 4.2. This helps in determining how many standard deviations a particular precipitation amount is from the mean.

The answer is indeed C.

point slope form is: y - y1=m(x - x1)

we substitute the values in; y1= 9; x1=1

and of course, the slope is 5

which gives us

y-9=5(x - 1)