What is the answer to the equation 7x-5y=20

What types of functions result from dividing two functions?

labwork [276]

Answer:

See how we can multiply or divide two functions to create a new function. Just like we can multiply and divide numbers, we can multiply and divide functions. For example, if we had functions f and g, we could create two new functions: f ⋅ g f\cdot g f⋅g and gfstart fraction, f, divided by, g, end fraction .

Step-by-step explanation:

6 0

3 years ago

Joseph has 3,134 pieces of paper in his classroom.emily has 5,229 pieces of paper in her classroom.how many pieces of paper do t

AysviL [449]

2095 is the correct answer

7 0

4 years ago

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Help please triangles and trig

Oksanka [162]

Answer:

$\sin L = \frac{3}{5}$

$\tan N = \frac{4}{3}$

$\cos L = \frac{4}{5}$

$\sin N = \frac{4}{5}$

Step-by-step explanation:

Given

The above triangle

First, we calculate the length LM using Pythagoras theorem.

$LN^2 = LM^2 + MN^2$

$10^2 = LM^2 + 6^2$

$100 = LM^2 + 36$

Collect like terms

$LM^2 = 100 - 36$

$LM^2 = 64$

Take positive square root

$LM=8$

Solving (a): Sin L

$\sin L = \frac{Opposite}{Hypotenuse}$

$\sin L = \frac{MN}{LN}$

$\sin L = \frac{6}{10}$

Simplify

$\sin L = \frac{3}{5}$

Solving (b): tan N

$\tan N = \frac{Opposite}{Adjacent}$

$\tan N = \frac{LM}{MN}$

$\tan N = \frac{8}{6}$

Simplify

$\tan N = \frac{4}{3}$

Solving (c): cos L

This calculated as:

$\cos L = \frac{Adjacent}{Hypotenuse}$

$\cos L = \frac{LM}{LN}$

$\cos L = \frac{8}{10}$

Simplify

$\cos L = \frac{4}{5}$

Solving (d): sin N

This is calculated using:

If $a + b = 90$

Then: $\sin a = \cos b$

So:

$\sin N = \cos L$

$\sin N = \frac{4}{5}$

8 0

3 years ago

The maximum walking speed S, in feet per second, of an animal can be modeled by the equation S=[gL, where g=32ft/sec and L is th

Artyom0805 [142]

Answer:

no clue

Step-by-step explanation:

i needed help w this too

3 0

3 years ago

A rocket is fired upward from some initial distance above the ground. Its height in feet, h,above the ground, t seconds after

kirill115 [55]

Check the picture below.

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h(t) = &{{ -16}}t^2&{{ +96}}t&{{ +2560}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so.. the maximum point is at $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}~~~ft$

and it happened after $\bf -\cfrac{{{ b}}}{2{{ a}}}~~~seconds$

--------------------------------------------------------------------------

now, when the rocket reaches the ground, as you saw on the picture, y = 0, thus

$\bf h(t)=-16t^2+96t+2560\implies 0=-16t^2+96t+2560
\\\\\\
0=-16(t^2-6t-160)\implies 0=t^2-6t-160
\\\\\\
0=(t-16)(t+10)\implies t=
\begin{cases}
\boxed{16}\\
-10
\end{cases}$

"t" is an amount in seconds, for this specific phenomena, it cannot be negative, so -10 is not a feasible value.

8 0

3 years ago

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