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marissa [1.9K]
3 years ago
8

If you substitute 12 + 2x for y in the second

Mathematics
2 answers:
sergeinik [125]3 years ago
8 0

Answer:

B

Step-by-step explanation:

the next 2 are

b

c and e

vladimir2022 [97]3 years ago
4 0
No second equation provided.
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What is the absolute value of the following complex number?<br> -4 + 4
snow_lady [41]

Answer:

Itz 0 if I am not wrong

hope it helps

3 0
3 years ago
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1. What is the volume of the prism that can be constructed from this net?
Irina-Kira [14]
1. V = lwh
    V = (8)(4)(5)
    V = (32)(5)
    V = 160 units³

2. V = lwh + lwh
    V = (4)(2)(1) + (3)(1)(1)
    V = (8)(1) + (3)(1)
    V = 8 + 3
    V = 11 units³

3. V = leh + s³
    V = (9)(3)(4) + (3)³
    V = (27)(4) + 27
    V = 108 + 27
    V = 135 units³

4. V = lwh
    V = (3)(7)(2)
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3 0
3 years ago
Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )
Zina [86]

Answer:

\purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

Step-by-step explanation:

\cos( \alpha ) =  \frac{ \sqrt{6} }{6}  =  \frac{1}{ \sqrt{6} }  \\  \\  \therefore \:  \sin( \alpha )  =  \sqrt{1 -  { \cos}^{2} ( \alpha ) }  \\  \\  =  \sqrt{1 -  \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} }  \\  \\ =  \sqrt{1 -  {\frac{1}{ {6} }}}  \\  \\ =  \sqrt{ {\frac{6 - 1}{ {6} }}}   \\  \\  \red{\sin( \alpha ) =  \sqrt{ { \frac{5}{ {6} }}} } \\  \\  \tan( \alpha ) =  \frac{\sin( \alpha ) }{\cos( \alpha ) }  =  \sqrt{5}  \\  \\ \sin( \beta )  =  \frac{ \sqrt{2} }{4}  \\  \\  \implies \: \cos( \beta )  =   \sqrt{ \frac{7}{8} }  \\  \\ \tan( \beta )  =  \frac{\sin( \beta ) }{\cos( \beta ) } =  \frac{1}{ \sqrt{7} }   \\  \\  \tan( \alpha  -  \beta ) =  \frac{ \tan \alpha  -  \tan \beta }{1 +  \tan \alpha .  \tan \beta}  \\  \\  =  \frac{ \sqrt{5} -  \frac{1}{ \sqrt{7} }  }{1 +  \sqrt{5} . \frac{1}{ \sqrt{7} } }  \\  \\  =  \frac{ \sqrt{35} - 1 }{ \sqrt{7}  +  \sqrt{5} }  \\  \\  \purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

8 0
3 years ago
Can someone please help me with these math problems ASAP?!<br> Thanks in advance! ;)
Mamont248 [21]

Answer:

11) Here given Function,

f(x) = \sqrt{2x-11}

And, g(x) = -|0.1|x+5

For f(x) = g(x)

\sqrt{2x-11}=-|0.1|x+5

2x-11=(-|0.1|x+5)^2

2x-11=0.01x^2 - 10|0.1|x+25

2x=0.01x^2 - 10|0.1|x+25+11

2x=0.01x^2 - 10|0.1|x+36

0.01x^2 - 10|0.1|x - 2x+36=0

When we solve this equation,

We found,

 x = 12.5227 ≈ 12.53

Thus, the required solution is, x = 12.53

12) Here the height of rocket A in x second,

f(x) = -16x^2+74x+9

And, The height gain by the rocket B in x seconds,

g(x) = -16x^2+82x

If at x seconds both A and B gain the same height,

That is, f(x) = g(x)

⇒ -16x^2+74x+9= -16x^2+82x

⇒74x + 9 = 82x

⇒ 82x - 74x = 9

⇒ 8x = 9

⇒ x = 1.125 ≈ 1.13

Thus, the required solution is x = 1.13 seconds (approx)


4 0
3 years ago
What is the coefficient of x2 in the expansion of (x + 2)3?
Vinvika [58]
I think 6 is the answer. Hopefully I answered your question

4 0
3 years ago
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