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3 years ago

Please help me expand theses and then help me factorise these brackets

Mathematics

2 answers:

-Dominant- [34]3 years ago

8 0

Answer:

-9xy+16x

Step-by-step explanation:

x(y-2)+3x(6-y)-7xy

xy-2x+18x-3xy-7xy

xy-3xy-7xy-2x+18x

-9xy+16x

melisa1 [442]3 years ago

3 0

Answer:

-9xy+16x

Step-by-step explanation:

x(y-2) +3x(6-y) -7xy

Distribute

xy -2x +18x-3xy -7xy

Combine like terms

-9xy+16x

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4^12*6^15*7^21<br><br> HEELPOOOOPO

Varvara68 [4.7K]

9514 1404 393

Answer:

4^4·6^5·7^7 = 1,639,390,814,208

Step-by-step explanation:

Taking the cube root multiplies each exponent by 1/3.

((4^12)(6^15)(7^21))^(1/3) = (4^(12/3))·(6^(15/3))·(7^(21/3)) = (4^4)(6^5)(7^7)

6 0

2 years ago

Choose the line that is the best fit for the data.

Svet_ta [14]

Answer:

I would say line 2

Step-by-step explanation:

just think it is

3 0

2 years ago

Use slopes and y-intercepts to determine if the lines 5x-5y=-2 and -x+2y=4

HACTEHA [7]

The given pair of lines are not perpendicular.

The line is a curve showing the shortest distance between 2 points.

5x - 5y = -2 - - - - - (1)
Transform the equation into standard form,
5x + 2 = 5y
y = 5x /5 + 2/5
y = x + 2/5

The slope of equation 1 is $m_1 = 1$ and intercept c = 2 / 5

Similarly
x + 2y = 4 - - - - - - - -(2)
Transform it into standard form
y = -x/2 + 4 /2
y = -x / 2 + 2

Slope of the equation 2 $m_2$ = -1 / 2 and intercept c = 2
Slope of line 1 * slope of line 2 = 1 * -1/2 = -1/2

Since the lines are not perpendicular because the pair of lines does not satisfy the property of perpendicular lines i.e
$m_1*m_2 = -1$

Thus, the given pair of lines are not perpendicular.

Learn more about lines here:

brainly.com/question/2696693

#SPJ1

6 0

1 year ago

Find the maximum value or minimum value for the function f(x) = 0.15(x + 1)² - 3.

il63 [147K]

Answer:

The minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

Step-by-step explanation:

Given function is $f(x) = 0.15(x + 1)^2 - 3$

We need to find the maximum value or the minimum value for the function.

Now, differentiate $f(x) = 0.15(x + 1)^2 - 3$ w.r.t $x$ .

$f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\$

$f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)$

Now, we will equate $f'(x)=0$ to find critical point.

$0.3(x+1)=0\\x=-1$

Plug this critical point in to the function $f(x) = 0.15(x + 1)^2 - 3$ we get,

$f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3$

Also, $f''(x)=0.3$ which is positive, We have minimum value.

So, the minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

7 0

3 years ago

PLEASE HELP ME

irina [24]

Answer: pt 2 HOPE THIS HELPS!

7 0

3 years ago

Read 2 more answers