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Bas_tet [7]
3 years ago
9

What equivalent to 5m(m+3)

Mathematics
1 answer:
Lostsunrise [7]3 years ago
3 0

Answer:

5m^2 + 15m

Step-by-step explanation:

use distributive property:

5m(m) is 5m^2

5m(3) is 15m

which is 5m^2 + 15m

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Who will answer first i will mark the brainliest
m_a_m_a [10]

Answer:

It is b

Step-by-step explanation:

7 0
2 years ago
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A mixture of compounds X and Y in a 0.100-cm cell had an absorbance of 0.215 at 272 nm and 0.191 at 327 nm. Find [X] and [Y] in
Oksi-84 [34.3K]

Answer: The concentration of X is 7.13582\times 10^{-6} and the concentration of Y is 2.53159\times 10^{-5}.

Step-by-step explanation:

Since we have given that

At 272 nm, absorbance = 0.215

At 327 nm, absorbance = 0.191

As we have given that

                               Compound X             Compound Y

272                               16400                            3870

327                                3990                             6420

So, our equations becomes

16400C_1+3870c_2=0.215\\\\3990C_1+6420C_2=0.191

By solving these two equations, we get that

C_1=7.13582\times 10^{-6}\\\\C_2=2.53159\times 10^{-5}

Hence, the concentration of X is 7.13582\times 10^{-6} and the concentration of Y is 2.53159\times 10^{-5}.

5 0
3 years ago
Using the unit circle find the exact value. CSC -30º =​
AlexFokin [52]

Answer:

csc (- 30)° = - 2

Step-by-step explanation:

Using the trigonometric relationship

csc x = \frac{1}{sinx} and the exact value sin30° = \frac{1}{2}

sin(- 30)° = - sin30° = - \frac{1}{2}, thus

csc (- 30)° = \frac{1}{-\frac{1}{2} } = - 2

8 0
3 years ago
The magnitude and direction of two vectors are shown in the diagram. What is the magnitude of their sum? ​
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Separate the vectors into their <em>x</em>- and <em>y</em>-components. Let <em>u</em> be the vector on the right and <em>v</em> the vector on the left, so that

<em>u</em> = 4 cos(45°) <em>x</em> + 4 sin(45°) <em>y</em>

<em>v</em> = 2 cos(135°) <em>x</em> + 2 sin(135°) <em>y</em>

where <em>x</em> and <em>y</em> denote the unit vectors in the <em>x</em> and <em>y</em> directions.

Then the sum is

<em>u</em> + <em>v</em> = (4 cos(45°) + 2 cos(135°)) <em>x</em> + (4 sin(45°) + 2 sin(135°)) <em>y</em>

and its magnitude is

||<em>u</em> + <em>v</em>|| = √((4 cos(45°) + 2 cos(135°))² + (4 sin(45°) + 2 sin(135°))²)

… = √(16 cos²(45°) + 16 cos(45°) cos(135°) + 4 cos²(135°) + 16 sin²(45°) + 16 sin(45°) sin(135°) + 4 sin²(135°))

… = √(16 (cos²(45°) + sin²(45°)) + 16 (cos(45°) cos(135°) + sin(45°) sin(135°)) + 4 (cos²(135°) + sin²(135°)))

… = √(16 + 16 cos(135° - 45°) + 4)

… = √(20 + 16 cos(90°))

… = √20 = 2√5

5 0
3 years ago
Read 2 more answers
Is anyone good at geometry???
zhenek [66]
Hey there Smarty!

This would be considered to be a (acute angel) which in this case, we would have to make sure that this whole triangle would equal less than 270 because each angle would be less than 90°

If we add/multiply \boxed{(90+90+90) \ or \ (90*3) = 270}.

So, we would have to know that was ever this would all add up to be, this would have to be less than 270°

\left[\begin{array}{ccc}\boxed{\boxed{(2.5+5) \\ \\ (2.5*2.5) \\ (2.5-2) \\ (2.5-1) \\}} \\ \\ this \ would \ be \ why \ I \ would \ say \\ that \ this \ would \ be \ the \ answer \end{array}\right]

I truly hope this helps, and also, it's kind of my 
 first time doing this I hope this would be helpful.
~Jurgen
7 0
3 years ago
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