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solniwko [45]
3 years ago
12

Find the corret measure of 6

Mathematics
2 answers:
antoniya [11.8K]3 years ago
4 0

if I'm correct, 3 would be 78 as well . I'm so sorry if I get this wrong . I struggled a bit with this part and math a lot , and I only worked on it for like 2 days and then I had to test for it

Degger [83]3 years ago
4 0

Answer:

∠3 = 102°

Step-by-step explanation:

║A straight line always equals 180°. So all we need to do is subtract 78° ║from 180° for the missing angle.

180 − 78 = 102

∠3 = 102°

Further explanation:

║The image shows two slanted lines across one vertical line. If we visualize ║a <em>circle</em> around the point where the slanted line intercepts with the vertical ║line, we can solve this.

<em>[we only need one line to work with, because both lines are parallel]</em>

║That circle you visualized divides the intercept into <em>four sectors</em>. One of ║those sectors, or angles, is labled 78°. We need to find the angle measure ║underneath the labled sector. The sector in the bottom right corner has ║the same angle measure as the labled sector, which is 78°. The sector ║labled ∠3 has the same angle measure as the sector in the upper right.

<em>[a full rotation around an angle is always 360°]</em>

║This can be easily solved by subtracting 78° from 180° to get the angle ║measure for the upper right sector, which is equivalent to ∠3. Using basic ║subtraction, 180° − 78° = 102°. We can prove this by multiplying 78° by 2, ║multiplying 102° by two, and adding the products.

78 ⋅ 2 = 156

102 ⋅ 2 = 204

204 + 156 = 360

║Because the sums equal 360°, we can determine the answer is true.

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Answer:

           \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0

Explanation:

Assuming the correct expression is to find the following limit:

         \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}

Use the property the limit of the quotient is the quotient of the limits:

         \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}

Evaluate the numerator:

          \frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}

Evaluate the denominator:

  • Since         \lim_{x \to1}sin(x-2)\neq 0

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3 years ago
Mr. Thomas drove 75 miles in May. He drove 6 times as many miles in July as he did in May. He
harkovskaia [24]

Answer:

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Step-by-step explanation:

No. of miles driven by Mr. Thomas in May = 75

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__________________________________________________

Another condition given   that miles driven in June is 4 times of miles driven by Mr. Thomas in July(450miles as calculated above).

Thus

No. of miles driven by Mr. Thomas in June = 4 * No. of miles driven by Mr. Thomas in July   = 4* 450 miles = 1800 miles.

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