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Papessa [141]
2 years ago
5

Multiply. 9i (5-2i)=

Mathematics
1 answer:
Vilka [71]2 years ago
6 0

Answer:

The answer is 45i + 18

since i *i=-1

You might be interested in
How many ways can you arrange 10 numbers
kkurt [141]

Answer:

3628800

Step-by-step explanation:

There are 10 options for the first number.

That leaves 9 options for the second number.

That leaves 8 options for the third number.

So on and so forth.

The number of ways 10 numbers can be arranged is:

10×9×8×7×6×5×4×3×2×1

= 10!

= 3628800

8 0
3 years ago
F(x) = 2x2 - 5<br> and<br> g(x) = 5x + 7<br> Find<br> f(g(x)) = [ ? ]x2 + [ ]x +
lbvjy [14]

Answer:10x^2 -18

Step-by-step explanation:since f(x) is becoming the new x then youd subtitute the equation into the x of the g(x) equation. which would be 5(2x^2 -5) +7 which is 10x^2 -25 +7.and you'd simplify from there.

8 0
2 years ago
I need Help please!!!
fomenos

Answer:

150 cm squared

Step-by-step explanation:

6 0
3 years ago
Two equations are given below:
astraxan [27]
A - 3b = 4
a = b -2

(b - 2) - 3b = 4
b - 2 - 3b = 4
-2b = 4 + 2
-2b = 6
b = 6/-2
b = -3

a = b - 2
a = -3 -2
a = -5

to check: a = -5 ; b = -3  
⇒ (-5,-3)
a - 3b = 4
-5 - 3(-3) = 4
-5 + 9 = 4
4 = 4
4 0
3 years ago
Please help I am so lost!!!
ASHA 777 [7]
\bf tan\left( \frac{x}{2} \right)+\cfrac{1}{tan\left( \frac{x}{2} \right)}\\\\&#10;-----------------------------\\\\&#10;tan\left(\cfrac{{{ \theta}}}{2}\right)=&#10;\begin{cases}&#10;\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}&#10;\\ \quad \\&#10;&#10;\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}&#10;\\ \quad \\&#10;&#10;\boxed{\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}}&#10;\end{cases}\\\\

\bf -----------------------------\\\\&#10;\cfrac{1-cos(x)}{sin(x)}+\cfrac{1}{\frac{1-cos(x)}{sin(x)}}\implies \cfrac{1-cos(x)}{sin(x)}+\cfrac{sin(x)}{1-cos(x)}&#10;\\\\\\&#10;\cfrac{[1-cos(x)]^2+sin^2(x)}{sin(x)[1-cos(x)]}\implies &#10;\cfrac{1-2cos(x)+\boxed{cos^2(x)+sin^2(x)}}{sin(x)[1-cos(x)]}&#10;\\\\\\&#10;\cfrac{1-2cos(x)+\boxed{1}}{sin(x)[1-cos(x)]}\implies \cfrac{2-2cos(x)}{sin(x)[1-cos(x)]}&#10;\\\\\\&#10;\cfrac{2[1-cos(x)]}{sin(x)[1-cos(x)]}\implies \cfrac{2}{sin(x)}\implies 2\cdot \cfrac{1}{sin(x)}\implies 2csc(x)
4 0
3 years ago
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