What is the answer to. Work out 5-10/-5

1/4 and 5/12 with common denominator

Zigmanuir [339]

Answer:If it's adding the answer will be 8/12 if it must be simplified =2/3

Step-by-step explanation:

First you time 1/4 by 3 and then you add.

6 0

2 years ago

Simplify. Your answer should contain only positive exponents with no fractional exponents in the denominator.

mart [117]

Answer:

$\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$ .

Step-by-step explanation:

The given expression is

$\dfrac{3y^{\frac{1}{4}}}{4x^{-\frac{2}{3}}y^{\frac{3}{2}}\cdot 3y^{\frac{1}{2}}}$

We need to simplify the expression such that answer should contain only positive exponents with no fractional exponents in the denominator.

Using properties of exponents, we get

$\dfrac{3}{4\cdot 3}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}}$ $[\because a^ma^n=a^{m+n}]$

$\dfrac{1}{4}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{2}}$

$\dfrac{1}{4}\cdot \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{y^{2}}$ $[\because a^{-n}=\dfrac{1}{a^n}]$

$\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$

We can not simplify further because on further simplification we get negative exponents in numerator or fractional exponents in the denominator.

Therefore, the required expression is $\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$ .

5 0

3 years ago

What is the positive slope of the asymptote of (y+11)^2/100-(x-6)^2/4=1? The positive slope of the asymptote is .

VladimirAG [237]

Answer:

the positive slope of the asymptote = 5

Step-by-step explanation:

Given that:

$\frac{(y+11)^2}{100} -\frac{(x-6)^2}{4} =1$

Using the standard form of the equation:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}= 1$

where:

(h,k) are the center of the hyperbola.

and the y term is in front of the x term indicating that the hyperbola opens up and down.

a = distance that indicates how far above and below of the center the vertices of the hyperbola are.

For the above standard equation; the equation for the asymptote is:

$y = \pm \frac{a}{b} (x-h)+k$

where;

$\frac{a}{b}$ is the slope

From above;

(h,k) = 11, 100

$a^2$ = 100

a = $\sqrt{100}$

a = 10

$b^2 =4$

b = $\sqrt{4}$

b = 2

$y = \pm \frac{10}{2} (x-11)+2$

$y = \pm 5 (x-11)+2$

y = 5x-53 , -5x -57

Since we are to find the positive slope of the asymptote: we have

$\frac{a}{b}$ to be the slope in the equation $y = \pm \frac{10}{2} (y-11)+2$

$\frac{a}{b}$ = $\frac{10}{2}$

$\frac{a}{b}$ = 5

Thus, the positive slope of the asymptote = 5

8 0

2 years ago

Andrew wants to measure the height of a traffic light. He walks exactly 20 feet from the base of the traffic light and looks up.

slava [35]

Answer-

The height of the traffic light is 21.8 feet.

Solution-

As shown in figure, we can see the right triangle formed from the situation,

From the question,

BC = the distance walked by Andrew from the traffic light = 20 feet

Height of Andrew's eye = 5 feet

Height of the traffic light = AB + Height of Andrew's eye

From the properties of triangle,

$\Rightarrow \tan \theta =\frac{Height}{Base}$

$\Rightarrow \tan C =\frac{AB}{BC}$

$\Rightarrow AB=\tan C\times BC$

$\Rightarrow AB=\tan 40\times 20$

$\Rightarrow AB=16.8$

Therefore, height of the traffic light = AB + Height of Andrew's eye = 16.8+5 = 21.8 feet

4 0

3 years ago

Read 2 more answers

If the parent function f(x) = |x| is transformed to g(x) = |x| + 4, what transformation occurs from f(x) to g(x)?

sergey [27]

Answer:

graph is shifted upward to create g(x)

6 0

2 years ago