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3 years ago

Which pair of angles are corresponding angles?

Mathematics

1 answer:

Effectus [21]3 years ago

7 0

Corresponding angles are angles that are the same, and occupy the same relative position.

1 and 2: Incorrect - These are supplementary angles

3 and 5: Incorrect - These are same-side interior angles

2 and 6: Correct - These are corresponding angles

4 and 7: Incorrect - These are alternate interior angles

Hope this helps!! :)

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Given the diagram below, find the value of x.

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Answer:

yes 11000

Step-by-step explanation:

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2 years ago

Please help I promise it is easy <br><br> algebra 1

Alborosie

Answer:

y=1/x-1/5

Step-by-step explanation:

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2 years ago

Find the work done by the force field F(x, y) = xi + (y + 5)j in moving an object along an arch of the cycloid r(t) = (t − sin(t

Nimfa-mama [501]

Integrate the force field along the given path (call it <em>C</em>):

$W=\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=\int_0^{2\pi}\mathbf F(x(t),y(t))\cdot\frac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}\bigg((t-\sin t)\,\mathbf i+(6-\cos t)\,\mathbf j\bigg)\cdot\bigg((1-\cos t)\,\mathbf i+\sin t\,\mathbf j\bigg)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}(t-t\cos t+5\sin t)\,\mathrm dt=\boxed{2\pi^2}$

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3 years ago

a teacher awarded 50 marks to his least student in a class test. How many marks was awarded to the best student if he was 40% be

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3 years ago

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73

faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]

Therefore,

CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})$

Thus,

Lower limit of CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}})$

Lower limit of CI = $(98.3 - \frac{2.62\times0.73}{\sqrt{103}})$

Lower limit of CI = 98.11

and,

Upper limit of CI = $(Mean + \frac{t\times\sigma}{\sqrt{n}})$

Upper limit of CI = $(98.3 + \frac{2.62\times0.73}{\sqrt{103}})$

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F

7 0

3 years ago