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3 years ago

A rectangle is 25 in. long and 13 in. wide what is the area? gives brainliest to right answer

Mathematics

2 answers:

Serggg [28]3 years ago

4 0

The area is 325 in ^2

monitta3 years ago

3 0

The area is 325 in ^2

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Suppose that y varies inversely with x. Use the information to find k, and then choose the equation of variation. x = 1.2 when y

baherus [9]

Y = k/x where k is the constant of variation
y = 3 , x = 1.2 gives
3 = k/1.2
therefore
k = 3*1.2 = 3.6
and the required equation is

y = 3.6 / x

6 0

3 years ago

Point K is on line JL. given JL =4x, JK= 2x+3, and KL=x, determine the numerical length of kl

balandron [24]

Hi! I'm happy to help!

Our total line is JL (4x), and it is split into two parts: JK, and KL. We have our values, and we know that JK+KL=JL, so we can substitute our values and solve for x:

4x=(2x+3)+(x)

4x=3x+3

To solve for x, we have to isolate it on one side of the equation.

First, let's subtract 3x from both sides so that we can isolate x:

4x=3x+3

-3x -3x

x=3

So, our x=3, which means that KL=3.

I hope this was helpful, keep learning! :D

5 0

2 years ago

2(x + 4) + 2x + 3 = 3x + 16 what does x eqaul

Ad libitum [116K]

Answer:

x=5

Step-by-step explanation:

2(x+4)+2x+3=3x+16

One solution was found :

x = 5

7 0

3 years ago

Read 2 more answers

If u is a unit vector, find u · v and u · w. (assume v and w are also unit vectors.) equilateral triangle

serg [7]

solution:

we know that ,

u.v = ΙuΙ ΙvΙcosθ

here,

θ =60° (since the given triangle is equilateral triangle)

u.v = ΙuΙ ΙvΙcos60°

= 1 x 1 x 1/2

u.v = 1/2

now, u.w = ΙuΙ ΙwΙcosθ

= ΙuΙ x cos(60x2)

u.w = -1/2

5 0

3 years ago

Determine which function has the greatest rate of change over the interval [0, 2].

ruslelena [56]

Remember that the average rate of change of a function over an interval is the slope of the straight line connecting the end points of the interval. To find those slopes, we are going to use the slope formula: $m= \frac{y_{2}-y_{1}}{x_2-x_1}$

Rate of change of $a$ :
From the graph we can infer that the end points are (0,1) and (2,4). So lets use our slope formula to find the rate of change of $a$ :
$m= \frac{y_{2}-y_{1}}{x_2-x_1}$
$m= \frac{4-1}{2-0}$
$m= \frac{3}{2}$
$m=1.5$
The average rate of change of the function $a$ over the interval [0,2] is 1.5

Rate of change of $b$ :
Here the end points are (0,0) and (2,2)
$m= \frac{2-0}{2-0}$
$m= \frac{2}{2}$
$m=1$
The average rate of change of the function $b$ over the interval [0,2] is 1

Rate of change of $c$ :
Here the end points are (0,-1) and (2,0)
$m= \frac{0-(-1)}{2-0}$
$m= \frac{1}{2}$
$m=0.5$
The average rate of change of the function $c$ over the interval [0,2] is 0.5

Rate of change of $d$ :
Here the end points are (0,0.5) and (2,2.5)
$m= \frac{2.5-0.5}{2-0}$
$m= \frac{2}{2}$
$m=1$
The average rate of change of the function $d$ over the interval [0,2] is 1

We can conclude that the function that has the greatest rate of change over the interval [0, 2] is the function a.

4 0

3 years ago