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3 years ago

X-1/(x+1)(x-5)

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

6 0

Answer:

x=1

Step-by-step explanation:

x-1/(x+1)(x-5)

When the numerator is equal to zero and the denominator is not equal to zero

x-1= 0

x=1

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What is the value of h(−2) when h(x)=3x 1? enter your answer in the box. i dont like just getting answers without trying to figu

marta [7]

I hope this helps you

x= -2 h (-2)= 3. (-2)+1= -6+1= -5

x=-2 h (-2)= 3. (-2) -1= -6-1= -7

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3 years ago

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Ilya [14]

Answer:

6 times 10 times 3.

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3 years ago

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Sphere A and Sphere B , are similar. The volumes of A and B, are 17 and 136 cubic centimetres, respectively. The diameter of B ,

xxMikexx [17]

Answer:

Step-by-step explanation:

Given: Sphere A and Sphere B are similar.

The volumes of A and B are 17 $cm^3$ and 136

The diameter of B is 6 cm.

To find: diameter of A

Solution:

Let R denotes radius of sphere A and r denotes radius of sphere B.

Radius of sphere A= R

Diameter of sphere B = 6 cm

So, radius of sphere B (r) = $\frac{6}{2}=3\,\,cm$

Volume of sphere is $\frac{4}{3}\pi(radius)^3$

Volume of sphere A = $\frac{4}{3}\pi(R)^3$

$\frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3}=\frac{17}{136}=\frac{1}{8}\\\frac{R^3}{r^3}=\frac{1}{8}\\\frac{R}{r}=\frac{1}{2}\\r=2R$

Put r = 3 cm

$3=2R\\R=\frac{3}{2}=1.5\,\,cm$

Diameter of sphere A = 2 × Diameter

= 2 × 1.5

=3 cm

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3 years ago

X^3+8 divided by x+2

Klio2033 [76]

Answer: x^2+4

Step-by-step explanation:

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10. Determine whether or not, vectors ui(1,-2, 0, 3), u2 = (2, 3,0,-1), u3 = (3,9,-4,-2) e R is a linear combination of the (2,-

vodka [1.7K]

If (2, -1, 2, 1) is a linear combination of the three given vectors, then there should exist $c_1,c_2,c_3$ such that

$(2,-1,2,1)=c_1(1,-2,0,3)+c_2(2,3,0,-1)+c_3(3,9,-4,-2)$

or equivalently, there should exist a solution to the system

$\begin{cases}c_1+2c_2+3c_3=2\\-2c_1+3c_2+9c_3=-1\\-4c_3=2\\3c_1-c_2-2c_3=1\end{cases}$

Right away we get $c_3=-\dfrac12$ , so the system reduces to

$\begin{cases}c_1+2c_2=\dfrac72\\\\-2c_1+3c_2=\dfrac72\\\\3c_1-c_2=0\end{cases}$

Notice that the first equation is the sum of the latter two. The third equation gives us

$3c_1-c_2=0\implies 3c_1=c_2$

so that in the second equation,

$-2c_1+3c_2=\dfrac72\implies7c_1=\dfrac72\implies c_1=\dfrac12$

which in turn gives

$3c_1=c_2\implies c_2=\dfrac32$

and hence the (2, -1, 2, 1) is a linear combination of the given vectors, with

$\boxed{(2,-1,2,1)=\dfrac12(1,-2,0,3)+\dfrac32(2,3,0,-1)-\dfrac12(3,9,-4,-2)}$

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4 years ago