All categories

2 years ago

Expand 8(r + 6) please help !

Mathematics

1 answer:

Levart [38]2 years ago

6 0

Answer:

8r + 48

Step-by-step explanation:

In order to expand brackets you have to multiply everything inside the brackets by the number on the outside.

Therefore:

8 x r = 8r

8x 6 = 48

Put these together and you end up with 8r + 48.

You might be interested in

Jeremy wants to borrow $100 from his brother. His brother agreed but

andrey2020 [161]

I think it's 12.50

Im not sure so sorry if im wrong but im 99% sure

7 0

2 years ago

Read 2 more answers

Which expression could be modeled by decimal grid?

faust18 [17]

Answer:

D) 0.80×0.40 so the answer is D

7 0

2 years ago

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte

Mashcka [7]

Answer:

f'(x) > 0 on $(-\infty ,\frac{1}{e})$ and f'(x)<0 on $(\frac{1}{e},\infty)$

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

$f'(x)>0$

To find its decreasing interval :

$f'(x)$

2) Then let's find the critical point of this function:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0$

2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x'' $=\frac{1}{e}$ ≈0.37 for e≈2.72

$x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37$

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0

2 years ago

What is the solution to this equation -4x + 2=26

wolverine [178]

$-4x+2=26$ <em>subtract 2 from both sides</em>

$-4x=24$ <em>divide both sides by (-4)</em>

$\boxed{x=-6}$

8 0

2 years ago

Read 2 more answers

One of the roots of the equation 3x^2+7x−q=0 is −5. Find the other root and q.

statuscvo [17]

Answer: The answer is $\textup{The other root is }\dfrac{8}{3}~\textup{and}q=40.Step-by-step explanation: The given quadratic equation is[tex]3x^2+7x-q=0\\\\\Rightarrow x^2-\dfrac{7}{3}x-\dfrac{q}{3}=0.$

Also given that -5 is one of the roots, we are to find the other root and the value of 'q'.

Let the other root of the equation be 'p'. So, we have

$p-5=-\dfrac{7}{3}\\\\\\\Rightarrow p=5-\dfrac{7}{3}\\\\\\\Rightarrow p=\dfrac{8}{3},$

and

$p\times(-5)=-\dfrac{q}{3}\\\\\\\Rightarrow \dfrac{8}{3}\times 5=\dfrac{q}{3}\\\\\\\Rightarrow q=40.$

Thus, the other root is $\dfrac{8}{3}$ and the value of 'q' is 40.

3 0

2 years ago