A. y:24=6:y Can someone please help me

Mathematics

1 answer:

Nikolay [14]3 years ago

6 0

Answer:

y = ± 12

Step-by-step explanation:

Given

y : 24 = 6 : y, or in fraction form

$\frac{y}{24}$ = $\frac{6}{y}$ ( cross- multiply )

y² = 144 ( take the square root of both sides )

y = ± $\sqrt{144}$ = ± 12

You might be interested in

(1,2) and (3,-8) in slope-intercept form

ycow [4]

Y=5x-3. First, you have to find the slope, which is 10/2 or 5/1. Then, you plug in the x and y with either of the 2 coordinates (I chose the first), and sole for the y-intercept.

8 0

3 years ago

What are the coordinates of the vertex of the function below write your answer in the form y-9= -6(x-1)^2

Oliga [24]

Answer:

The co-ordinates of the vertex of the function y-9= -6(x-1)^2 is (1, 9)

Solution:

Given, equation is $y-9=-6(x-1)^{2}$

We have to find the vertex of the given equation.

When we observe the equation, it is a parabolic equation,

We know that, general form of a parabolic equation is $y-9=-6(x-1)^{2}$

Where, h and k are x, y co ordinates of the vertex of the parabola.

$\text { Now, parabola equation is } y-9=-6(x-1)^{2} \rightarrow y=-6(x-1)^{2}+9$

By comparing the above equation with general form of the parabola, we can conclude that,

a = -6, h = 1 and k = 9

Hence, the vertex of the parabola is (1, 9).

7 0

2 years ago

Will someone help with this ! I was allowed a retry for it and I don’t know how to do it

pashok25 [27]

Answer:

$\frac{dN}{dt}=k(725-N)\\separating the variables and integrating\\\int \frac{dN}{725-N}=\int kdt+c\\-log (725-N)=kt+c\\log (725-N)=-kt-c\\(725-N)=e^{-kt-c} =e^{-kt} *e^{-c} =Ce^{-kt} \\when t=0,N=400\\725-400=Ce^{0} =C\\C=325\\when t=3,N=650\\725-650=325e^{-3k} \\\frac{75}{325}=(e^{-k})^3\\\\e^{-k} =(\frac{3}{13}) ^{\frac{1}{3} } \\when t=5\\725-N=325(\frac{3}{13}) ^{\frac{5}{3} } =325*\frac{3}{13}*(\frac{3}{13} )^{\frac{2}{3}}\\N=725-75*(\frac{3}{13} )^{\frac{2}{3} }=725-28=697$