All categories

4 years ago

2nq=3+7nq solve for n thanks

Mathematics

2 answers:

Gnesinka [82]4 years ago

7 0

Answer:

n = - $\frac{3}{5q}$

Step-by-step explanation:

Given

2nq = 3 + 7nq ( subtract 7nq from both sides )

- 5nq = 3 ( divide both sides by the multiplier - 5q )

n = $\frac{3}{-5q}$ = - $\frac{3}{5q}$

Nataly_w [17]4 years ago

5 0

Answer:

n= -3/q

Step-by-step explanation:

2nq-7nq=3

-5nq=3

n = -3/q

You might be interested in

(2a^-2)(3a^3b^2)(c^-2)

aalyn [17]

Answer:

Step-by-step explanation:

7 0

3 years ago

A rectangle has a height of 2 + 9 and a width of x2 + 2x.

Dafna1 [17]

Answer: 40

Step-by-step explanation:

8 0

3 years ago

If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c

horsena [70]

Answer:

Part 1) $cos(A + B) = \frac{140}{221}$

Part 2) $cos(A - B) = \frac{153}{185}$

Part 3) $cos(A - B) = \frac{84}{85}$

Part 4) $cos(A + B) = -\frac{36}{85}$

Part 5) $cos(A - B) = \frac{63}{65}$

Part 6) $cos(A+ B) = -\frac{57}{185}$

Step-by-step explanation:

<u><em>the complete answer in the attached document</em></u>

Part 1) we have

$sin(A)=\frac{8}{17}$

$cos(B)=\frac{12}{13}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{8}{17})^2=1$

$cos^2(A)+\frac{64}{289}=1$

$cos^2(A)=1-\frac{64}{289}$

$cos^2(A)=\frac{225}{289}$

$cos(A)=\pm\frac{15}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{15}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{13})^2=1$

$sin^2(B)+\frac{144}{169}=1$

$sin^2(B)=1-\frac{144}{169}$

$sin^2(B)=\frac{25}{169}$

$sin(B)=\pm\frac{25}{169}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{5}{13}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}$

$cos(A + B) = \frac{180}{221}-\frac{40}{221}$

$cos(A + B) = \frac{140}{221}$

Part 2) we have

$sin(A)=\frac{3}{5}$

$cos(B)=\frac{12}{37}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{3}{5})^2=1$

$cos^2(A)+\frac{9}{25}=1$

$cos^2(A)=1-\frac{9}{25}$

$cos^2(A)=\frac{16}{25}$

$cos(A)=\pm\frac{4}{5}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{4}{5}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{37})^2=1$

$sin^2(B)+\frac{144}{1,369}=1$

$sin^2(B)=1-\frac{144}{1,369}$

$sin^2(B)=\frac{1,225}{1,369}$

$sin(B)=\pm\frac{35}{37}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{35}{37}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}$

$cos(A - B) = \frac{48}{185}+\frac{105}{185}$

$cos(A - B) = \frac{153}{185}$

Part 3) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}$

$cos(A - B) = \frac{24}{85}+\frac{60}{85}$

$cos(A - B) = \frac{84}{85}$

Part 4) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}$

$cos(A + B) = \frac{24}{85}-\frac{60}{85}$

$cos(A + B) = -\frac{36}{85}$

Download odt

4 0

4 years ago

Please help! you'll get Brainliest if you do!

alisha [4.7K]

Answer:

Step-by-step explanation:

3 0

3 years ago

It takes Max three hours to run 30 kilometers. Georges, on the other hand, travels the same distance in 2 hours and 20 minutes.

Mamont248 [21]

George will arrive 20 minutes before Max

It takes Max 3 hours(180 minutes) to run 30 km .

The rate can be calculated as follows:

rate = 30 / 180 = 1/6 km/ min

George uses 2 hours 20 minutes(140 minutes) to run 30 km.

The rate can be calculated as follows:

rate = 30 / 140 = 3/14 km / min

If both are to complete a 15-kilometre race, Therefore,

Max time will be

1 / 6 = 15 / t

t = 15 × 6

t = 90 minutes

George time will be

3/14 = 15 / t

3t = 210

t = 210 / 3

t = 70 minutes

Time difference = 90 - 70 = 20 minutes

Therefore, George will arrive 20 minutes earlier than Max

3 0

3 years ago