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VladimirAG [237]
3 years ago
11

Please hurry I will mark you brainliest 8(3 - i) = 5(2 - 2i)

Mathematics
1 answer:
Mrac [35]3 years ago
6 0

Answer:-7

Step-by-step explanation:

24-8i=10-10i

24-10=-10i+8i

14=-2i

I=-7

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Answer:

6 cups of sugar left

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7 0
2 years ago
5(6x - 1) - 10x + 5 = 4(x + 10) - 4x
Dmitry [639]

Answer:

x=2

Step-by-step explanation:

5(6x-1)-10x+5=4(x+10)-4x

simplify both sides

30x-5-10x+5=4x+40-4x

Simplify even further

20x=40

x=2

3 0
3 years ago
Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions
quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of v_0 and F_2 be the component normal to v_0.

Since, |v_0| = 1,

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, v_0 = \left

From figure,

|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3

We know that the direction of F_1 is opposite of the direction of v_0, so we have

$F_1 = -5\sqrt3 v_0$

    $=-5\sqrt3 \left$

    $= \left$

The unit vector in the direction normal to the plane, v_1 has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5

∴  F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>

                   $=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

           $= = F$

3 0
2 years ago
Work out the answers of the following:
schepotkina [342]

Answer:

1)-5

2)-15

3)1

4)-11

5)-70

if negative use () for example

(-2)-7+4

7 0
3 years ago
Read 2 more answers
the sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger
Reil [10]
<h2>Answer:</h2>

x\geq 3 \ and \ x+1 \geq 4

<h2>Step-by-step explanation:</h2>

The question in this problem is:

<em>The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?</em>

<em />

First of all, let's name the first variable x which is the smaller number. Accordingly, the lager number will be x+1 given that those numbers are consecutive. On the other hand<em> at most </em>conveys the idea of an inequality, which is:

\leq \\ which \ means \ less \ than

So:

1. The sum of 2 consecutive integers can be written as:

v+(v+1)

2. Nine times the smaller and 5 times the larger can be written as:

9v-5(v+1)

Finally, the whole statement:

The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger:

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\ \frac{6}{2} \leq \frac{2x}{2} \\ \\ 3 \leq x \\ \\ x\geq 3 \\ \\ and \\ \\ x+1 \geq 4

The two numbers are:

x\geq 3 \ and \ x+1 \geq 4

6 0
3 years ago
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