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3 years ago

Wich expression represents the phrase one fifth the sum of 14 and a number

Mathematics

1 answer:

kompoz [17]3 years ago

7 0

Answer:

(14+a)/5

(14+a) * 1/5

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What is the area of the triangular base?

Ludmilka [50]

Answer:

90cm2

Step-by-step explanation:

1/2x base x height.

1/2 x10 x 18

90cm2

4 0

2 years ago

What is the answer to 4+4-(5-1)

Zanzabum

Your answer would be 4 you go with P.E.M.D.A.S so (5-1)= 4 then 4+4=8-4=4

6 0

3 years ago

Triangle ABC is congruent to triangle DFE. Find x.

Archy [21]

ΔABC ≡ ΔDFE

⇒ line EF = BC

<u>Solve x:</u>

3x - 4 = 20

3x = 24

x = 8

Answer: x = 8

4 0

3 years ago

Solve the following equation:

Rama09 [41]

Complete the square.

$z^4 + z^2 - i\sqrt 3 = \left(z^2 + \dfrac12\right)^2 - \dfrac14 - i\sqrt3 = 0$

$\left(z^2 + \dfrac12\right)^2 = \dfrac{1 + 4\sqrt3\,i}4$

Use de Moivre's theorem to compute the square roots of the right side.

$w = \dfrac{1 + 4\sqrt3\,i}4 = \dfrac74 \exp\left(i \tan^{-1}(4\sqrt3)\right)$

$\implies w^{1/2} = \pm \dfrac{\sqrt7}2 \exp\left(\dfrac i2 \tan^{-1}(4\sqrt3)\right) = \pm \dfrac{2+\sqrt3\,i}2$

Now, taking square roots on both sides, we have

$z^2 + \dfrac12 = \pm w^{1/2}$

$z^2 = \dfrac{1+\sqrt3\,i}2 \text{ or } z^2 = -\dfrac{3+\sqrt3\,i}2$

Use de Moivre's theorem again to take square roots on both sides.

$w_1 = \dfrac{1+\sqrt3\,i}2 = \exp\left(i\dfrac\pi3\right)$

$\implies z = {w_1}^{1/2} = \pm \exp\left(i\dfrac\pi6\right) = \boxed{\pm \dfrac{\sqrt3 + i}2}$

$w_2 = -\dfrac{3+\sqrt3\,i}2 = \sqrt3 \, \exp\left(-i \dfrac{5\pi}6\right)$

$\implies z = {w_2}^{1/2} = \boxed{\pm \sqrt[4]{3} \, \exp\left(-i\dfrac{5\pi}{12}\right)}$

3 0

1 year ago

Solve for x. Show your work.<br> -1/2x <-12

vitfil [10]

Answer:

$x>24$

Step-by-step explanation:

First, let's re-write it multiplied by -2 on each side, the denominator will disappear and we'll have our x positive:

$x>24$

Remember: if you multiply a inequality for a negative number, change the sign to its opposite, for example the < become the >

5 0

2 years ago

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