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butalik [34]
3 years ago
5

15/9 =25/9x I simplified it as much as I could but i'm stuck here:/

Mathematics
2 answers:
anyanavicka [17]3 years ago
5 0
The Simplest Version would
be 5/3 if that’s the answer you’re looking for :)
kvv77 [185]3 years ago
4 0

Answer:

Step-by-step explanation:

We move all terms to the left: 15/9-(25/9x)=0 Domain of the equation: 9x)!=0 x!=0/1 x!=0 x∈R

We add all the numbers together, and all the variables

-(+25/9x)+15/9=0

We get rid of parentheses

-25/9x+15/9=0

We calculate fractions

=0 x=0/1 x=0

i hope helps you and if you want you can give me a brainly crown only if you want

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A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro
pshichka [43]

Answer:

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed \frac{dx}{dt} = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}

(h)\frac{dh}{dt}=(x)\frac{dx}{dt}

(h)\frac{dh}{dt}=rx

\frac{dh}{dt}=\frac{rx}{h}

When automobile is 30 miles past the intersection,

For x = 30

\frac{dh}{dt}=\frac{30r}{h}

Since h=\sqrt{d^{2}+(30)^{2}}

Therefore,

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

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