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Lerok [7]
3 years ago
7

What does 5/4x = 1/2

Mathematics
1 answer:
aleksklad [387]3 years ago
7 0
You would multiply 4 on both sides
5x=2
then you would divide 5 on both sides
x=2/5
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Radda [10]

Answer:

TU and TS (with arrows on top)

Step-by-step explanation:

The sides of any angle are rays. This are half lines, which in your case both start at the point T, and one goes towards point U, while the other one goes from toward point S. So I would write TU (with a little arrow symbol on top). and in the other box I would write TS also with a little arrow symbol on top.

5 0
3 years ago
Of fabric costs $5.00 per yard, how much will 7/8 yards cost?
Maksim231197 [3]

Answer: It would be $4.38



Step-by-step explanation: If you divide the $5 by .8 it will give you what 1/8 of it would be then you multiply it by 7 to get the 7/8 of it.


4 0
3 years ago
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lina2011 [118]

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2 years ago
How many gold medals have been awarded for the 1500-meter race from 1896 - 2000 ?
Nataly [62]

Answer:

Hola disculpa no me acuerdo

Step-by-step explanation:

De verdad que me encantaria ayudar pero no me acuerdo :(

8 0
1 year ago
Read 2 more answers
HELP!! Algebra help!! Will give stars thank u so much <333
Anna35 [415]

Answers:

  • Part a)  \bf{\sqrt{x^2+(x^2-3)^2}
  • Part b)  3
  • Part c)   2.24
  • Part d)  1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying d = \sqrt{x^2 + (x^2-3)^2} is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0
2 years ago
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