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1 year ago

−5x 5 y 5 (−6xy 3 ) simplify

Mathematics

2 answers:

Readme [11.4K]1 year ago

8 0

Answer:

|-2| + 2

Step-by-step explanation:

-4 / 2-5i

cos (x) / 1 - sin squared (x)

|-2| + 2

KengaRu [80]1 year ago

3 0

Answer:

Step-by-step explanation:

<h2>Expression multiplication:</h2>

$\sf -5x^5*y^5 *(-6xy^3) = (-5)*(-6) *x^{5+1}*y^{(5+3)}\\$

$\sf = 30x^6y^8$

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Find the inverse of the function. y=x2 + 4x + 4

adoni [48]

Answer: To find the inverse of the function, we need to make x as a function of y and at the final step make a switch between x and y (i.e. make x as y and y as x)

y = x² + 4x + 4 ⇒⇒⇒ factor the quadratic equation

y = (x+2)(x+2)

y = (x+2)² ⇒⇒⇒ take the square root to both sides

√y = x+2

x = √y - 2 ⇒⇒⇒ x becomes a function of y

final step:

∴ y = √x - 2 ⇒⇒⇒ the inverse of the given function

So, as a conclusion:

f(x) = y = x² + 4x + 4 ⇒⇒⇒ the given function

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3 years ago

Three pounds of lawn seed covers 1800 square feet. How many 4 pound bags are needed to cover 8400 square feet?

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2 years ago

Find the area of a triangle bounded by the y-axis, the line f(x)=9−4/7x, and the line perpendicular to f(x) that passes through

Setler79 [48]

ANSWER:

The area of the triangle bounded by the y-axis is $\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

SOLUTION:

Given, $f(x)=9-\frac{-4}{7} x$

Consider f(x) = y. Hence we get

$f(x)=9-\frac{-4}{7} x$ --- eqn 1

$y=9-\frac{4}{7} x$

On rewriting the terms we get

4x + 7y – 63 = 0

As the triangle is bounded by two perpendicular lines, it is an right angle triangle with y-axis as hypotenuse.

Area of right angle triangle = $\frac{1}{ab}$ where a, b are lengths of sides other than hypotenuse.

So, we need find length of f(x) and its perpendicular line.

First let us find perpendicular line equation.

Slope of f(x) = $\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{7}$

So, slope of perpendicular line = $\frac{-1}{\text {slope of } f(x)}=\frac{7}{4}$

Perpendicular line is passing through origin(0,0).So by using point slope formula,

$y-y_{1}=m\left(x-x_{1}\right)$

Where m is the slope and $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$

$y-0=\frac{7}{4}(x-0)$

$y=\frac{7}{4} x$ --- eqn 2

4y = 7x

7x – 4y = 0

now, let us find the vertices of triangle, one of them is origin, second one is point of intersection of y-axis and f(x)

for points on y-axis x will be zero, to get y value, put x =0 int f(x)

0 + 7y – 63 = 0

7y = 63

y = 9

Hence, the point of intersection is (0, 9)

Third vertex is point of intersection of f(x) and its perpendicular line.

So, solve (1) and (2)

$\begin{array}{l}{9-\frac{4}{7} x=\frac{7}{4} x} \\\\ {9 \times 4-\frac{4 \times 4}{7} x=7 x} \\\\ {36 \times 7-16 x=7 \times 7 x} \\\\ {252-16 x=49 x} \\\\ {49 x+16 x=252} \\\\ {65 x=252} \\\\ {x=\frac{252}{65}}\end{array}$

Put x value in (2)

$\begin{array}{l}{y=\frac{7}{4} \times \frac{252}{65}} \\\\ {y=\frac{441}{65}}\end{array}$

So, the point of intersection is $\left(\frac{252}{65}, \frac{441}{65}\right)$

Length of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right)$ and (0,9)

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(9-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+0} \\ &=\frac{252}{65} \end{aligned}$

Now, length of perpendicular of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right) \text { and }(0,0)$

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(0-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+\left(\frac{441}{65}\right)^{2}} \\ &=\frac{\sqrt{(12 \times 21)^{2}+(21 \times 21)^{2}}}{65} \\ &=\frac{63}{65} \sqrt{65} \end{aligned}$