Find a coterminal angle between 0° and 360°. 780

1 answer:

3 0

for an angle of 780°, we can say that is really a 360° + 360° + 60°, so two full revolutions plus an extra 60°. Check the picture below, with the coterminal in green.

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I'll mark brainly pls do this

lidiya [134]

Step-by-step explanation:

Verbs are read, complete, checked, work, invite. Pronouns are their, your, and them. Nouns are Classroom, Zeroes, Friday, February, friends, assignment, students, trimester. Adjectives are simple, missing.

Hope this helps! If I missed some or got some wrong, please tell me! Goodbye!

8 0

3 years ago

Read 2 more answers

From 145 pounds to 132 pounds

Mariana [72]

From 145 pounds to 132 pounds

145 - 132 = 13

13 pounds

Check:-
132 + 13 = 145
Correct!

13 pounds

4 0

3 years ago

Solve -np-80<60 for n. Show your work. Solve 2a-5d=30 for d. Show your work.

xz_007 [3.2K]

$-np-80 < 60\qquad\text{change the signs}\\\\np+80$

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$2a-5d=30\qquad\text{subtract 2a from both sides}\\\\-5d=-2a+30\qquad\text{change the signs}\\\\5d=2a-30\qquad\text{divide both sides by 5}\\\\d=0.4a-6$

8 0

3 years ago

PLEASE ANSWER + BRAINLIEST!! Factor completely. 4k - 20k^9 = 3b^2 - 108 =

Neporo4naja [7]

$4k-20k^9=4k(1-5k^8)=4k\left[1^2-(k^4\sqrt5)^2\right]\\\\=4k(1-k^4\sqrt5)(1+k^4\sqrt5)=4k\left[1^2-(k^2\sqrt[4]5)^2\right](1+k^4\sqrt5)\\\\=4k(1-k^2\sqrt[4]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k\left[1^2-(k\sqrt[8]5)^2\right](1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k(1-k\sqrt[8]5)(1+k\sqrt[8]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)$

$3b^2-108=3(b^2-36)=3(b^2-6^2)=3(b-6)(b+6)$

$Used:\ (a-b)(a+b)=a^2-b^2$

7 0

3 years ago

Write the cosine ratio of angle b and show work please

Dennis_Churaev [7]

The formula for cosine is adj/hyp. For angle B, 10 is the adjacent and 24 is the opposite. The hypotenuse ( in this case, 26) is the longest side. the ratio for angle b would be 10/26.

7 0

3 years ago