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2 years ago

Question 21

Mathematics

1 answer:

horsena [70]2 years ago

7 0

Answer:

Step-by-step explanation:

A(x) = 4x ( x + 10 ) - 2x ( 8 ) + $\frac{2x(x)}{2}$ + $\frac{2x(x+2)}{2}$

A(x) = 4x² + x² + x² + 40x - 16x + 2x

<em>A(x) = 6x² + 26x</em>

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Solve this equation -k/5=7/10

liberstina [14]

$-\dfrac{k}{5}=\dfrac{7}{10}\\
-10k=35\\
k=-3.5$

4 0

3 years ago

Are greater than 0 and located to the right of 0 on a number

jolli1 [7]

Answer:

1. 0

2. true

hope this helps :)

3 0

3 years ago

Read 2 more answers

The width of a rectangle measures (2.5u+9.8)(2.5u+9.8) centimeters, and its length measures (1.5u+3.9)(1.5u+3.9) centimeters. Wh

icang [17]

Solution,

We have,

Width of rectangle, b = (2.5u+9.8) cm

Length of rectangle, l = (1.5u+3.9) cm

We need to find the expression for the perimeter of the rectangle. The formula for the perimeter of a rectangle is given by:

Perimeter = 2(l+b)

P = 2[(2.5u+9.8)+(1.5u+3.9)]

Collecting like terms

P = 2[(2.5u+1.5u)+(9.8+3.9)]

P=2(4u+13.7)

⇒ P = 8u+27.4

So, the expression for the perimeter of the rectangle is (8u+27.4) cm.

3 0

2 years ago

Given z1 = -3

trasher [3.6K]

Answer:

See explanation

Step-by-step explanation:

The given complex number are:

$z_1=-3\sqrt{3}+3i$

and

$z_2=6\cos 150\degree+6i\sin 150\degree$

When we rewrite $z_1=-3\sqrt{3}+3i$ in complex form, we obtain;

$z_1=r(\cos \theta+i\sin \theta)$

where

$r=\sqrt{(-3\sqrt{3})^2+3^2 }=\sqrt{36}=6$

and

$\theta=tan^{-1}(\frac{y}{x})$

$\implies \theta=tan^{-1}(\frac{-3\sqrt{3}}{3})=150\degree$

Hence,

$z_1=6(\cos 150\degree+i\sin 150\degree)$

$z_1=6\cos 150\degree+6i\sin 150\degree$

Hence

$z_1=z_2$

8 0

3 years ago

Find all constants a such that the vectors (a, 4) and (2, 5) are parallel.

Nezavi [6.7K]

Answer:

$\displaystyle a = \frac{8}{5}$ .

Step-by-step explanation:

Two vectors $(x_1,\, y_1)$ and $(x_2,\, y_2)$ are parallel to one another if and only if the ratio between their corresponding components are equal:

$\displaystyle \frac{x_1}{x_2} = \frac{y_1}{y_2}$ .

Equivalently:

$\displaystyle \frac{x_1}{y_1} = \frac{x_2}{y_2}$ .

For the two vectors in this equation to be parallel to one another:

$\displaystyle \frac{a}{4} = \frac{2}{5}$ .

Solve for $a$ :

$\displaystyle a = \frac{8}{5}$ .

$\displaystyle \frac{8}{5}$ would be the only valid value of $a$ ; no other value would satisfy the $\displaystyle \frac{x_1}{y_1} = \frac{x_2}{y_2}$ equation.

4 0

3 years ago