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3 years ago

Solve 5x + 14 = k for x.

Mathematics

2 answers:

inessss [21]3 years ago

7 0

Answer:

k - 14/5

Step-by-step explanation:

5x + 14 = k

On subtracting 14 on both sides,

5x = k - 14

On dividing by 5 on both sides,

x = k - 14/5

Therefore, the value of x = k-14/5

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Y_Kistochka [10]3 years ago

4 0

Answer:

x = (k-14)/5

Step-by-step explanation:

5x + 14 = k

Subtract 14 from each side

5x + 14-14 = k-14

5x = k-14

Divide by 5

5x/5 = (k-14)/5

x = (k-14)/5

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Which pair of equations can be used to solve for x in PQR?

almond37 [142]

Answer:

Step-by-step explanation:

sin57° = $\frac{opposite}{hypotenuse}$ = $\frac{PQ}{PR}$ = $\frac{x}{18}$

∠ P = 180° - (90 + 57)° ← sum of angles in triangle

∠ P = 180° - 147° = 33° , then

cos33° = $\frac{adjacent}{hypotenuse}$ = $\frac{PQ}{PR}$ = $\frac{x}{18}$

Thus the 2 ratios used to solve for x are A

7 0

3 years ago

Students sold yogurt to raise money for graduation. The first 20 yogurts sold cost $5 each. To sell more yogurt they lowered the

goblinko [34]

5(20) + 4y = 216
100 + 4y = 216
4y = 216 - 100
4y = 116
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7 0

3 years ago

Show that a = −1 + √3i and b = 2 satisfy 1/a+b=1/a + 1/b

Zarrin [17]

Answer:

LHS = $\frac{1 - \sqrt3i}{4}$ = RHS = $\frac{1 - \sqrt3i}{4}$

Step-by-step explanation:

Data provided in the question:

a = −1 + √3i and b = 2

to prove:

$\frac{1}{a+b}=\frac{1}{a} + \frac{1}{b}$

Considering the LHS

⇒ $\frac{1}{a+b}$

substituting the value of a and b, we get

⇒ $\frac{1}{−1 + \sqrt3i+2}$

⇒ $\frac{1}{1 + \sqrt3i}$

on multiplying and dividing by conjugate ( 1 - √3i )

we get

$\frac{1}{1 + \sqrt3i}\times\frac{1 - \sqrt3i}{1 - \sqrt3i}$

$\frac{1 - \sqrt3i}{(1^2 - (\sqrt3i)^2}$

$\frac{1 - \sqrt3i}{1 + 3}$ (as (√i)² = -1 )

$\frac{1 - \sqrt3i}{4}$

Now,

considering the RHS

$\frac{1}{a} + \frac{1}{b}$

substituting the value of a and b, we get

⇒ $\frac{1}{-1 + \sqrt3i} + \frac{1}{2}$

⇒ $\frac{2\times1 + ( -1 + \sqrt3i)\times1}{(-1 + \sqrt3i)\times2}$

⇒ $\frac{2 + ( -1 + \sqrt3i)}{(-1 + \sqrt3i)\times2}$

⇒ $\frac{1 + \sqrt3i}{(-1 + \sqrt3i)\times2}$

now,

on multiplying and dividing by conjugate ( -1 - √3i )

we get

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1 - \sqrt3i}{-1 - \sqrt3i}$

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1( 1 + \sqrt3i)}{-1 - \sqrt3i}$

$\frac{(1 + \sqrt3i}^2\times(-1){((-1)^2 - (\sqrt3i)^2)\times2}$

$\frac{(1^2 + (\sqrt3i)^2+2(1)(\sqrt3i)\times(-1)}{(1 + 3)\times2}$

$\frac{(1 - 3 + 2\sqrt3i)\times(-1)}{(4)\times2}$

$\frac{(-2 + 2\sqrt3i)\times(-1)}{(4)\times2}$

$\frac{-2( 1 - 2\sqrt3i)\times(-1)}{(4)\times2}$

$\frac{( 1 - 2\sqrt3i)}{(4)}$

Since, LHS = RHS

hence satisfied

3 0

3 years ago

Missing Terms- ?, 21, 16, ?, 6, ?

Svet_ta [14]

Answer: 26, 21, 16, 11, 6, 1 (You're going up by 5 each time)

Step-by-step explanation: Good luck! :D

7 0

4 years ago

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victus00 [196]

Answer:

$2\sqrt{6}$

Step-by-step explanation:

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5 0

3 years ago

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