How would I put this in standard form?
1 answer:
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<h3>Two answers: 5, 7</h3>
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Explanation:
A drawing may be helpful to see what's going on. Check out the diagram below. This is one way of drawing out the two triangles. The locations of the points don't really matter, and neither does the the orientation of how you rotate things. What does matter is we have the right points connected to form the segments mentioned.
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For now, focus on triangle TIP only. In order to have this be isosceles, we must make TP = 5 or TP = 7.
If TP = 5, then it's the same length as TI.
If TP = 7, then it's the same length as PI.
In either case, we have exactly two sides the same length (the other side different) which is what it means for a triangle to be isosceles.
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Let's consider triangle TOP. For it to be isosceles, we must have two sides the same length. We already locked in TP to be either 5 or 7 in the previous section above. So there's no way that TP could be 11 units long to match up with PO = 11.
If TP = 5, then OT must also be 5 units long so that triangle TOP is isosceles.
If TP = 7, then OT = 7 for similar reasoning.
Either way, TP only has two choices on what it could be.
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In short, we basically just write the first two values given to us to get the two triangles to be isosceles. We can't use TP = 11 as it would make triangle TIP to be scalene (all sides are different lengths).
Answer:
y = - 24
Step-by-step explanation:
Given that y varies directly with x then the equation relating them is
y = kx ← k is the constant of variation
To find k use the condition y = 12 when x = - 3
k = =
= - 4
y = - 4x ← equation of variation
When x = 6, then
y = - 4 × 6 = - 24
Answer:
Our decision rule will be to reject The null hypothesis H0 if the test statistic is less than -1.645, or if it is greater than +1.645.
Step-by-step explanation:
The hypotheses would be;
Null hypothesis;H0: μ = 500
Alternative hypothesis;Ha: μ ≠ 500
Since it's two - tailed at 0.1 level of significance, then each tail will contain 5% or 0.05. From the z-table attached, the corresponding critical value of 0.05 is approximately 1.645 standard deviations from the mean.
Thus, our decision rule will be to reject The null hypothesis H0 if the test statistic is less than -1.645, or if it is greater than +1.645.
