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Keith_Richards [23]
3 years ago
13

The day my sibling reads a book I did, is the day I will never be allowed to read again. Iykyk

Mathematics
2 answers:
Darya [45]3 years ago
7 0

Answer:

Ok what is the question?

Rainbow [258]3 years ago
4 0

Answer:

yes they do allowed the book

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Given that 3 x : 10 = 5 : 2 Calculate the value of x .
borishaifa [10]

\tt{ 3x:10=5:2 } ⠀

\tt{\dfrac{3x}{10}=\dfrac{5}{2}  } ⠀

\tt{3x×2=10×5  } ⠀

\tt{6x=50  } ⠀

\tt{x=\dfrac{50}{6}  } ⠀

\tt{ x=\dfrac{25}{3}} ⠀

7 0
3 years ago
Which expression is equivalent to 6(4(6x-4y)-5) ?
iren2701 [21]

Answer:

144x - 96y - 30

Step-by-step explanation:

<u>Step 1:  Distribute </u>

6(4(6x - 4y) - 5)

6((4 * 6x) + (4 * -4y) - 5)

6(24x - 16y - 5)

<u>Step 2:  Distribute again </u>

<u />(6 * 24x) + (6 * -16y) + (6 * -5)

144x - 96y - 30

Answer:  144x - 96y - 30

8 0
3 years ago
Jason solved the following equation to find the value for x.
Ainat [17]
Plug in 6.5 for x.
-8.5x-3.5x=-78
-8.5(6.5)-3.5(6.5)=-78
-55.25-22.75=-78
-78=-78
5 0
1 year ago
6.) Alan, Becky, Jesus, and Mariah are four students in the chess club. If two of these students will be selected to represent t
Artist 52 [7]

Let's begin by listing out the information given to us:

There are four students: n = 4

Number of students to be selected: r = 2

To calculate the combination of 2 students to be chosen, we use:

\begin{gathered} ^4C_2=\frac{n!}{(n-r)!}=\frac{4!}{(4-2)!}=\frac{4!}{2!} \\ ^4C_2=\frac{4\cdot3\cdot2\cdot1}{2\cdot1}=\frac{4\cdot3}{1}=12 \\ ^4C_2=12 \end{gathered}

Therefore, there 12 possible combinations from these

3 0
1 year ago
A system for tracking ships indicated that a ship lies on a hyperbolic path described by 5x2 - y2 = 20. the process is repeated
zysi [14]
Answer:
The ship is located at (3,5)

Explanation:
In the first test, the equation of the position was:
5x² - y² = 20 ...........> equation I
In the second test, the equation of the position was:
y² - 2x² = 7 ..............> equation II
This equation can be rewritten as:
y² = 2x² + 7 ............> equation III

Since the ship did not move in the duration between the two tests, therefore, the position of the ship is the same in the two tests which means that:
equation I = equation II

To get the position of the ship, we will simply need to solve equation I and equation II simultaneously and get their solution.

Substitute with equation III in equation I to solve for x as follows:
5x²-y² = 20
5x² - (2x²+7) = 20
5x² - 2y² - 7 = 20
3x² = 27
x² = 9
x = <span>± </span>√9

We are given that the ship lies in the first quadrant. This means that both its x and y coordinates are positive. This means that:
x = √9 = 3

Substitute with x in equation III to get y as follows:
y² = 2x² + 7
y² = 2(3)² + 7
y = 18 + 7
y = 25
y = +√25
y = 5

Based on the above, the position of the ship is (3,5).

Hope this helps :)
8 0
3 years ago
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