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guajiro [1.7K]
3 years ago
6

Jodie ran at an average speed of 6 km/h for 10 minutes. How far did she run in km?

Mathematics
1 answer:
Leokris [45]3 years ago
5 0

Answer:

Distance cover by Jodie = 1 kilometer

Step-by-step explanation:

Given;

Speed of Jodie = 6 kilometer/ hour

Total time Jodie ran = 10 minutes

Find:

Distance cover by Jodie

Computation:

Total time Jodie ran = 10 minutes = 10 / 60 = 0.1667 hour

Distance = Speed x time

Distance cover by Jodie = Speed of Jodie x Total time Jodie ran

Distance cover by Jodie = 6 x 0.1667

Distance cover by Jodie = 1.002

Distance cover by Jodie = 1 kilometer

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Which graph represents the function f (x) = StartFraction 2 Over x minus 1 EndFraction + 4?
Pepsi [2]

Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

f(x) = \dfrac{2}{x - 1} + 4

From the given function, we have;

When x = 1, the denominator of the fraction, \dfrac{2}{x - 1}, which is (x - 1) = 0, and the function becomes, \dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as <em>x</em> increases, the fraction \dfrac{2}{x - 1} tends to 0, therefore as x increases, we have;

\lim_  {x \to \infty}  \dfrac{2}{(x - 1)} \to 0, and \  \dfrac{2}{(x - 1)}  + 4 \to 4

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, \dfrac{2}{0 - 1} + 4 = 2. When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

5 0
3 years ago
Point M is the midpoint of segment QR. If QM = 16 + x and MR = 2(x + 2), find the length of QM.
pav-90 [236]
Just add them
QM+MR=QR
16+x+2(x+2)=QR
16+x+2x+4=QR
20+3x=QR
6 0
3 years ago
Solve the expression below, when x = 8 <br> (x - 4) + 12
LenaWriter [7]

Answer:

16

Step-by-step explanation:

Lets solve this step by step,

First we insert 8 in x's place:

(8 - 4) + 12

Now we solve:

8 - 4 = 4

(4) + 12

4 + 12 = 16

Answer: 16

Hope this Helps

3 0
3 years ago
This is correct? I need help please :(
Leto [7]

Answer:

well, we only no that

2l + 2w is 80

(lengths and widths on all 4 sides

and that w * l is A

so the function should be w times what's left for l, but also expressed by something with w

2l + 2w = 80 | -2w

2l = 80 -2w | devide by 2

l = 80 - w

now we can substitute l in

A = w * l

so that we only need w's

A(w) = w * (80-w)

for any w it will give us the area, makes only sense for 0<w<80

8 0
3 years ago
Read 2 more answers
I desperately need help !
Lorico [155]

Answer:

t = 9.57

Step-by-step explanation:

We can use trig functions to solve for the t

Recall the 3 main trig ratios

Sin = opposite / hypotenuse

Cos = adjacent / hypotenuse

Tan = opposite / adjacent.

( note hypotenuse = longest side , opposite = side opposite of angle and adjacent = other side )

We are given an angle as well as its opposite side length ( which has a measure of 18 ) and we need to find its adjacent "t"

When dealing with the opposite and adjacent we use trig ratio tan.

Tan = opp / adj

angle measure = 62 , opposite side length = 18 and adjacent = t

Tan(62) = 18/t

we now solve for t

Tan(62) = 18/t

multiply both sides by t

Tan(62)t = 18

divide both sides by tan(62)

t = 18/tan(62)

t = 9.57

And we are done!

3 0
3 years ago
Read 2 more answers
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