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2 years ago

Please help! will give brainliest:)

Mathematics

1 answer:

Lorico [155]2 years ago

7 0

Answer

1) x = 8

2) x = 1/3y + -5/3z

3) x = -1/2y + -3/4z + -1/2

Hope this helps:))

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Work out the surface area of this solid prism.

alexandr1967 [171]

Answer:

2520cm²

Step-by-step explanation:

Surface area = sum of the area of all the bases.

Area of bottom base = 28 * 30 = 840

Area of right base = 25 * 30 = 750

Area of left base = 17 * 30 = 510

Area of two triangles = ( 15 * 28 ) / 2 = 210 * 2 = 420

Then add all the areas together to find surface area

Surface area = 840 + 750 + 510 + 420 = 2520cm²

Relevant Formulas:

Area of a rectangle = length * width. This formula was used to find the area of the bottom base, the right base and the left base

Area of a triangle = ( height * base length ) / 2 . This formula was used to find the area of the sides which are triangles.

5 0

2 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

Please help i will give extra points

klemol [59]

Answer:

$36 \div (3 \times 2) - 4 \\ 36 \div 6 - 4 \\ 6 - 4 \\ = 2$

<h3>PLEASE MARK ME BRAINLIEST</h3>

5 0

2 years ago

Suppose a spherical balloon grows in such a way that after t seconds, its volume is V = 4 sqrt(t) cm3. What is the volume of the

Arisa [49]

: You need to know the derivative of the sqrt function. Remember that sqrt(x) = x^(1/2), and that (d x^a)/(dx) = a x^(a-1). So (d sqrt(x))/(dx) = (d x^(1/2))/(dx) = (1/2) x^((1/2)-1) = (1/2) x^(-1/2) = 1/(2 x^(1/2)) = 1/(2 sqrt(x)).

There is a subtle shift in meaning in the use of t. If you say "after t seconds", t is a dimensionless quantity, such as 169. Also in the formula V = 4 sqrt(t) cm3, t is apparently dimensionless. But if you say "t = 169 seconds", t has dimension time, measured in the unit of seconds, and also expressing speed of change of V as (dV)/(dt) presupposes that t has dimension time. But you can't mix formulas in which t is dimensionless with formulas in which t is dimensioned.

Below I treat t as being dimensionless. So where t is supposed to stand for time I write "t seconds" instead of just "t".

Then (dV)/(d(t seconds)) = (d 4 sqrt(t))/(dt) cm3/s = 4 (d sqrt(t))/(dt) cm3/s = 4 / (2 sqrt(t)) cm3/s = 2 / (sqrt(t)) cm3/s.

Plugging in t = 169 gives 2/13 cm3/s.

4 0

3 years ago

The drama club spent $608 on food for a party for its 38 members. Let a be the amount spent on food per person.

serg [7]

Part A is C
part B is 16

5 0

3 years ago