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Tamiku [17]
3 years ago
8

How can I do this exercise?

%202%7D" id="TexFormula1" title="\frac{-2}{12} = \frac{x + 1}{x + 2}" alt="\frac{-2}{12} = \frac{x + 1}{x + 2}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
matrenka [14]3 years ago
3 0

Answer:

-8/7 =x

Step-by-step explanation:

-2         x+1

----- = -----------

12         x+2

Using cross products

-2 (x+2) = 12 (x+1)

Distribute

-2x -4 = 12x+12

Add 2x to each side

-2x+2x -4 = 12x+2x+12

-4 = 14x +12

Subtract 12 from each side

-4-12 = 14x

-16 = 14x

Divide each side by 14

-16/14 =14x/14

-8/7 =x

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Point R is on line segment QS. Given QR = 2 and RS=10, determine the length
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Answer:

Assuming you are talking about segment QS, it's 12 units.

Step-by-step explanation:

10+2=12.

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A right triangle has a hypotenuse of 10 mi. and a leg of 6 mi., what is the length of the other leg?
slava [35]

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8mi

Step-by-step explanation:

Using Pythagoras theorem

Hypotenuse square = opposite square + adjacent square.

Let the other leg be x

10 square = 6 square + x square

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Collect like terms

X square = 100 -36

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I hope this was helpful, Please mark as brainliest

7 0
3 years ago
Determine whether each point lies on the graph of the equation.(a) 2x - y - 3 = 0 (a) (1, 2)Yes, the point is on the graph.No, t
Neko [114]

Answer:

a) (1, 2) not on the graph.

b) (1, -1) is on the graph.

Step-by-step explanation:

Given the equation of the line as:

2x - y - 3 = 0

The points given are

a) (1, 2)

To determine whether it is on the graph of the line or not.

To do so, we can do 2 things:

1. Draw the graph and plot the point on the graph to check whether it is on the graph or not.

2. To put the point in the given equation of the line, whether the equation is satisfied or not.

For method 1: Kindly refer to the attached image of the line and point plotted.

Method 2:

Let us put x=1, y=2 in the Left Hand Side (LHS) of equation.

2\times 1 - 2 - 3\\\Rightarrow -3 \neq 0 , RHS

(1, 2) Not on the graph.

(b) (1, -1)

For method 1: Kindly refer to the attached image of the line and point plotted.

Method 2:

Let us put x=1, y=-1 in the Left Hand Side (LHS) of equation.

2\times 1 - (-1) - 3\\\Rightarrow 3-3 = 0 = RHS

(1, -1) is on the graph.

3 0
4 years ago
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