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2 years ago

1/4 equals how many eighths

Mathematics

1 answer:

alexira [117]2 years ago

6 0

Answer:

The answer = 0.25

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Need help with this math

alexira [117]

A hole occurs when both numerator and denominator of a rational function have the same factor.

Step-by-step explanation:

While graphing rational function, it has to be converted into the lowest terms by factoring the numerator and denominator. If the numerator and denominator has the same factor, a hole is said to have occurred and to solve the rational function, you have to set the common factor to zero.

After you set it to zero and solve, you obtain the x value which can be then used to find out the value of y.

8 0

2 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

A certain kind of bacteria multiplies at a rate of 150 percent per hour. When a person’s blood has 45,000,000 of these bacteria,

DIA [1.3K]

Initial number of Bacteria = Ao = 100
Number of Bacteria which make the person ill = A = 45,000,000

Growth rate = r = 150% = 1.50

The equation model for this situation is:

$A= A_{o}(1+r)^{t}$

Using the values in above equation, we can write:

$45000000=100(1+1.5)^{t} \\ \\ 
450000=(2.5)^{t} \\ \\ 
log(450000)=log((2.5)^{t}) \\ \\ 
log(450000)=t*log(2.5) \\ \\ 
t= \frac{log(450000)}{log(2.5)} \\ \\ 
t=14.21$

This means, the person will fall sick after about 14 hours.

8 0

3 years ago

Read 2 more answers

2. Which variable will eliminate? * 3x +6y=27 -3x+8y= 1

Anarel [89]

9x=27
x=3
———
-3*3+8y=1
5\4

5 0

3 years ago

Read 2 more answers

Need help simplifying this

diamong [38]

The simplified answer is $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$ .

Step-by-step explanation:

$$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term $(\frac{3x+y}{3x+y})$ and 2nd term by $(\frac{z+2x}{z+2x})$

= $\frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= $\frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}$

Now, the denominators are same, you can subtract it.

= $\frac{\left(6 x^{2}+6 x^{2}+11 x y-4 x y-2 y z-2 y z+3 x z-6 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

= $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

Thus the simplified solution is $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

4 0

3 years ago