Find the value of 7x+3y when x=12 and y=-6

Mathematics

2 answers:

miv72 [106K]3 years ago

7 0

Answer:

Step-by-step explanation:

To get the values we just have to substitute the number of the value

7x + 3y = 7(12) + 3(-6)

84 - 18 = 66

kenny6666 [7]3 years ago

6 0

Answer:

Step-by-step explanation:

7 x + 3 = 7 (12) + 3 (-6)

84 - 18 = 66

Substitute the value of the variable into the expression and simplify.

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One quarter of a certain number added to one eighth of the same number is 12. What is the number

alexdok [17]

Answer:

Step-by-step explanation:

let the number be n , then

$\frac{1}{4}$ n + $\frac{1}{8}$ n = 12 ( multiply through by 8 to clear the fractions )

2n + n = 96

3n = 96 ( divide both sides by 3 )

n = 32

The number is 32

6 0

2 years ago

What is the measure of NRS?

never [62]

Answer:

Step-by-step explanation:

(7x - 1) + 79° = 21x + 8

• An exterior angle of a triangle is equal to the sum of its two interior opposite angles.

$(7x - 1) +79 = 21x + 8 \\ 7x - 1 + 79 = 21x + 8 \\ 7x + 78 = 21x + 8 \\ 78 - 8 = 21x - 7x \\ 70 = 14x \\ \frac{70}{14} = x \\ 5 = x \\$

NRS = 21x + 8

$21x + 8 \\ 21(5) + 8 \\ 105 + 8 \\ {113}^{o}$

6 0

2 years ago

Please help with this problem.

larisa86 [58]

Answer:

60°, 120°

Step-by-step explanation:

$\frac{ {tan}^{2}x }{2} - 2 {cos}^{2}x = 1 \\ \\ \frac{ {tan}^{2}x - 4{cos}^{2}x }{2} = 1 \\ \\ {tan}^{2}x - 4{cos}^{2}x = 2 \\ \\ \frac{{sin}^{2}x}{{cos}^{2}x} - 4{cos}^{2}x = 2 \\ \\ \frac{{sin}^{2}x - 4{cos}^{4}x}{{cos}^{2}x} = 2 \\ \\ {sin}^{2}x - 4{cos}^{4}x = 2{cos}^{2}x \\ \\ 4{cos}^{4}x + 2{cos}^{2}x - {sin}^{2}x = 0 \\ \\ 4{cos}^{4}x + 2{cos}^{2}x + {cos}^{2}x - 1= 0 \\ \\ 4{cos}^{4}x + 3{cos}^{2}x - 1= 0 \\ \\ 4{cos}^{4}x + 4{cos}^{2}x - {cos}^{2}x - 1= 0 \\ \\4{cos}^{2}x({cos}^{2}x + 1) - 1({cos}^{2}x + 1) = 0 \\ \\ ({cos}^{2}x + 1)(4{cos}^{2}x - 1) = 0 \\ \\ ({cos}^{2}x + 1) = 0 \: or \: (4{cos}^{2}x - 1) = 0 \\ \\ {cos}^{2}x = - 1 \: or \: 4{cos}^{2}x = 1 \\ \\ {cos}x = \sqrt{ - 1} \: which \: is \: not \: possible \\ \therefore \: {cos}^{2}x = \frac{1}{4} \\ \\ \therefore \: {cos}x = \pm\frac{1}{2} \\ \\ \therefore \: {cos}x = \frac{1}{2} \: or \: {cos}x = - \frac{1}{2} \\ \\ \therefore \: {cos}x = {cos}60 \degree \: or \: {cos}x = {cos}120 \degree \\ \\ \therefore \:x = 60 \degree \: \: or \: \: x = 120 \degree$