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Naddik [55]
4 years ago
12

Solve for x algebraically:

Mathematics
1 answer:
Masja [62]4 years ago
5 0
<span>THE GIVEN QUESTION; 7x -3(4x-8) < 6x +12 - 9x THE GIVEN INTERVAL OF X IS FROM [4,8] SO THE VALUE OF X ={4,5,6,7,8} THE GIVEN QUESTION CAN BE EVALUATED TO 7x - 12x+ 24 < 12 - 3x -5x + 24 < 12 - 3x //USING BODMAS RULE NOW SUBSTITUTING THE VALUE OF X: 1) X=4 SOL: 4<0 2) X=5 SOL: -1<-3 3) X=6 SOL: -6<-6 4) X=7 SOL: -11<-9 5) X=8 SOL: -16<-12 HENCE THE VALUE OF X=7,8 SATISFY THE GIVEN INEQUALITY</span>
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A tiene la mitad de lo que tiene B. Si A gana 66 y B pierde 90. A tendrá del doble de lo que le quede a B ¿cuanto tiene cada uno
Oksana_A [137]

Respuesta:

a = 82; b 164

Explicación paso a paso:

a = 1 / 2b - - - (1)

a gana 66 = a + 66

b = b - 90

a = 2b

a + 66 = 2 (segundo - 90) - - - (2)

Ponga a = 1 / 2b en (2)

1 / 2b + 66 = 2 (b-90)

0.5b + 66 = 2b - 180

0.5b - 2b = - 180 - 66

-1,5b = - 246

b = 164

a = 1 / 2b

a = 1/2 * 164

a = 82

7 0
3 years ago
Suppose that A and B are nonsingular matrices. Then AB is also nonsingular. Furthermore, a theorem from linear algebra then stat
kherson [118]

Answer:

A) Verified

B) Proved

Step-by-step explanation:

a) Let's verify it for 2 x 2 matrix,

A=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] and B=\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]

AB=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]=\left[\begin{array}{ccc}a.e+b.g&a.f+b.h\\c.e+d.g&c.f+d.h\end{array}\right]

(AB)^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]

A^{-1}=\frac{1}{a.d-b.c} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right]

B^{-1}=\frac{1}{e.h-f.g} \left[\begin{array}{ccc}h&-f\\-g&e\end{array}\right]

B^{-1}A^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]

So it is proved that the results are same.

b) Now, let's prove it for any n x n matrix.

(AB)(AB)^{-1}=I\\\\A^{-1}(AB)(AB)^{-1}=A^{-1}I\\\\IB(AB)^{=1}=A^{-1}I\\\\B(AB)^{=1}=A^{-1}\\\\B^{-1}B(AB)^{=1}=B^{-1}A^{-1}\\\\I(AB)^{=1}=B^{-1}A^{-1}\\\\(AB)^{=1}=B^{-1}A^{-1}

8 0
3 years ago
Jebrel borrowed $7000 from his friend. He saves $110 each month to pay him back.
forsale [732]

Answer:

What is the question? This is just a statement

Step-by-step explanation:

5 0
3 years ago
 determine the base, b, of the exponential model. Is the base a growth or decay factor? a. b is 1.0394; It is a growth factor. b
Fed [463]
The answer is A.) b is 1.0349 it is a growth factor.
5 0
3 years ago
What is the interquartile range of the following data set?<br> 290, 357, 380, 413, 395, 313, 335
kolezko [41]

Answer:

123

Step-by-step explanation:

4 0
3 years ago
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