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2 years ago

Equations definitions

Mathematics

1 answer:

astraxan [27]2 years ago

4 0

Answer:

Step-by-step explanation:

MEAN - a). Finds the mean of a numerical data set.

MIN - f). Finds the lowest value of a data set.

COUNT - c). Counts the number of values input for a highlighted data set

MEDIAN - d). Finds the middle value of a numerical data set

MAX - e). Finds the greatest value for a data set

SUM - g). Adds all the numbers in a highlighted range together

MODE - b). Finds the number that occurs most often in a highlighted data set.

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'calcula que signo tiene el producto de cien números enteros, sabiendo que 27 de ellos son positivos y que ninguno es cero''.

prisoha [69]

Answer:

si a

Step-by-step explanation:

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2 years ago

1. Simplify the expression. Assume that

ololo11 [35]

Answer:

b. 6|x|

Step-by-step explanation:

√(36x²) = √36√(x²)= 6|x|

Because √ from the number can be only positive.

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2 years ago

For the function f(x) = x + 5, what is the ordered pair for the point on the graph when x = 4w? (1 point)

Dominik [7]

We are given the function f(x) = x + 5 in which the abscissa chosen is at x = 4w. To find the ordinate or the y-component, we replace x with 4w in the equation given. In this case, y = 4w + 5. Hence the answer to this problem is B. (4w, 4w + 5)

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2 years ago

Read 2 more answers

A person, who decided to go to weekend trip should not exceed 8 hours driving in a day. Average speed of forward journey is 60 k

Paraphin [41]

Since he must not exceed 8 hours driving in a day:
Let the distance of the picnic be = x km.

Therefore time for forward journey = x / 60
Return journey    = x / 50

The total trip should not exceed 8 hours.
Therefore: x / 60 + x / 50 <= 8.    LCM = 300

Taking LCM and multiplying on both sides:
5x + 6x <= 8(300)
   11x <= 2400
   x <= 2400/11
   x   <= 218.18
The picnic spot must be less than or equal to 218.18 km.

5 0

2 years ago

SOLVE THE QUESTION BELOW ASAP

qwelly [4]

Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

Let

h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

we have

$h(t)=-16t^{2}+98$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts, and the axis of symmetry

<em>Find the vertex</em>

The function is written in vertex form

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98$

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

$-16t^{2}+98=0$

$t^{2}=\frac{98}{16}$

square root both sides

$t=\pm\frac{\sqrt{98}}{4}$

$t=\pm7\frac{\sqrt{2}}{4}$

therefore

The t-intercepts are

$(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)$

$(-2.475,0), (2.475,0)$

Find the axis of symmetry

The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

$x=0$ ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point $(0,98)$

The y-intercept is the point $(0,98)$

The t-intercepts are $(-2.475,0), (2.475,0)$

The axis of symmetry is the y-axis

The graph in the attached figure

Part B) How far is the artifact fallen from the time t=0 to time t=1

we know that

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98\ ft$

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

Find the difference

$98\ ft-82\ ft=16\ ft$

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

For t=2

$h(t)=-16(2)^{2}+98$

$h(2)=34\ ft$

Find the difference

$82\ ft-34\ ft=48\ ft$

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

therefore

The distance traveled from t=1 to t=2 is greater than the distance traveled from t=0 to t=1

8 0

2 years ago