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s2008m [1.1K]
2 years ago
13

Given: Y is the midpoint of XZ Prove: XY = 1/2 XZ Statements Reasons

Mathematics
1 answer:
malfutka [58]2 years ago
7 0

Answer:

The proposition <em>If Y is the midpoint of XZ, then XY = 1/2 XZ </em>is true.

Step-by-step explanation:

We proceed to present the demonstration of the following proposition:

<em>If Y is the midpoint of XZ, then XY = 1/2 XZ.</em>

1) Y is the midpoint of XZ. Given.

2) XY + YZ = XZ Definition of line segment.

3) XY = YZ By 1)

4) 2\cdot XY = XZ 3) in 2)

5) XY = \frac{1}{2}\cdot XZ Algebra/Result

Therefore, the proposition <em>If Y is the midpoint of XZ, then XY = 1/2 XZ </em>is true.

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Help me on this question! Jake can carry 6 1/4 pounds of wood in from the barn.His father can carry 1 5/7 times as much as jake.
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\text{ Jake father can carry } 10\frac{5}{7} \text{ or } \frac{75}{7} \text{ pounds }

<em><u>Solution:</u></em>

From given question,

Number of pounds Jake carry = 6\frac{1}{4} \text{ pounds }

Number of pounds his father carry is 1\frac{5}{7} times as much as jake

To find: Number of pounds Jake father can carry

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\rightarrow 6\frac{1}{4} = \frac{4 \times 6+1}{4} = \frac{25}{4}\\\\\rightarrow 1\frac{5}{7} = \frac{7 \times 1+5}{7} = \frac{12}{7}

<em><u>Then according to question,</u></em>

\text{Pounds father carry } = \frac{12}{7} \text{ times the pounds jake carry }\\\\\text{Pounds father carry } = \frac{12}{7} \times \frac{25}{4}\\\\\text{Pounds father carry } = \frac{75}{7}\\\\\text{In terms of mixed fractions, }\\\\\text{Pounds father carry } = \frac{75}{7} = 10\frac{5}{7}

\text{ Thus, Jake father can carry } 10\frac{5}{7} \text{ or } \frac{75}{7} \text{ pounds }

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