Help please!! (ASAP)

What is the value of x in the equation 3x – one-ninthy = 18, when y = 27?

coldgirl [10]

Answer:

7

Step-by-step explanation:

3x- 1/9y = 18

value of x when y is 27

3x- 1/9*27 = 18

3x- 3 = 18

3x = 18 + 3

3x = 21

x= 21/3

x = 7

Hope this helps you!

4 0

3 years ago

Read 2 more answers

For what values of the variable, do the following fractions exist: y^2-1/y+y/y-3

ale4655 [162]

Answer:

Remember that the division by zero is not defined, this is the criteria that we will use in this case.

1) $\frac{y^2 - 1}{y} + \frac{y}{y - 3}$

So the fractions are defined such that the denominator is never zero.

For the first fraction, the denominator is zero when y = 0

and for the second fraction, the denominator is zero when y = 3

Then the fractions exist for all real values except for y = 0 or y = 3

we can write this as:

R / {0} U { 3}

(the set of all real numbers except the elements 0 and 3)

2) $\frac{b + 4}{b^2 + 7}$

Let's see the values of b such that the denominator is zero:

b^2 + 7 = 0

b^2 = -7

b = √-7

This is a complex value, assuming that b can only be a real number, there is no value of b such that the denominator is zero, then the fraction is defined for every real number.

The allowed values are R, the set of all real numbers.

3) $\frac{a}{a*(a - 1) - 1}$

Again, we need to find the value of a such that the denominator is zero.

a*(a - 1) - 1 = a^2 - a - 1

So we need to solve:

a^2 - a - 1 = 0

We can use the Bhaskara's formula, the two values of a are given by:

$a = \frac{-(-1) \pm \sqrt{(-1)^2 + 4*1*(-1)} }{2*1} = \frac{1 \pm \sqrt{5} }{2}$

Then the two values of a that are not allowed are:

a = (1 + √5)/2

a = (1 - √5)/2

Then the allowed values of a are:

R / {(1 + √5)/2} U {(1 - √5)/2}

7 0

2 years ago

Find the number of 3-digit numbers formed using the digits 1 to 9, without repetition, such the numbers either have all digits l

Tasya [4]

Answer: 120

Step-by-step explanation:

The total number of digits from 1 to 9 = 10

The number of digits from less than 5 (0,1,2,3,4)=5

Since repetition is not allowed so we use Permutations , then the number of 3-digit different codes will be formed :-

$^5P_3=\dfrac{5!}{(5-3)!}=\dfrac{5\times4\times3\times2!}{2!}=5\times4\times3=60$

The number of digits from greater than 4 (5,6,7,8,9)=5

Similarly, Number of 3-digit different codes will be formed :-

$^5P_3=60$

Hence, the required number of 3-digit different codes = 60+60=120

3 0

3 years ago

Find the equation of the tangent line to the curve (a lemniscate)

olya-2409 [2.1K]

Answer:

$m=\frac{9}{13}$ and $b=\frac{40}{13}$

Step-by-step explanation:

The equation of curve is

$2(x^2+y^2)^2=25(x^2-y^2)$

We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

$2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})$

$4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})$

The point of tangency is (-3,1). It means the slope of tangent is $\frac{dy}{dx}_{(-3,1)}$ .

Substitute x=-3 and y=1 in the above equation.

$4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})$

$40(-6+2\frac{dy}{dx})=25(-6-2\frac{dy}{dx})$

$-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}$

$80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240$

$130\frac{dy}{dx}=90$

Divide both sides by 130.

$\frac{dy}{dx}=\frac{9}{13}$

If a line passes through a points $(x_1,y_1)$ with slope m, then the point slope form of the line is

$y-y_1=m(x-x_1)$

The slope of tangent line is $\frac{9}{13}$ and it passes through the point (-3,1). So, the equation of tangent is

$y-1=\frac{9}{13}(x-(-3))$

$y-1=\frac{9}{13}(x)+\frac{27}{13}$

Add 1 on both sides.

$y=\frac{9}{13}(x)+\frac{27}{13}+1$

$y=\frac{9}{13}(x)+\frac{40}{13}$

Therefore, $m=\frac{9}{13}$ and $b=\frac{40}{13}$ .

5 0

3 years ago

What is the rate of change?

Ulleksa [173]

sheesh these are easy its 4

4 0

3 years ago