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2 years ago

13

Can someone help me? It's Algebra I. If you answer this can you explain how you did it?

1 answer:

Aloiza [94]2 years ago

3 0

I agree it is one and I used this calculator called MathPapa that can help with algebra you should try it. It is free.

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I would really appreciate it if someone could help me with these last few problems

lorasvet [3.4K]

Answer:

Ouestion 5 answers

1)8.5

2)150

3)8

4)15

Step-by-step explanation:

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1 year ago

Find the quotient of 1.2 divided by 251.2 show your work

Natali [406]

Do 1.2 then the division box and inside the box put 251.2 and the answer is <span>209.333333333 with an infinite 3</span>

4 0

2 years ago

Read 2 more answers

the value of c guaranteed to exist by the Mean Value Theorem for V(x) = x² in the interval [0, 3] is...? a) 1 b) 2 c) 3/2 d) 1/2

lilavasa [31]

Answer: c) $\dfrac{3}{2}$ .

Step-by-step explanation:

Mean value theorem : If f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that

$\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a} \cdot\end{displaymath}$

Given function : $f(x) = x^2$

Interval : [0,3]

Then, by the mean value theorem, there is at least one number c in the interval (0,3) such that

$f'(c)=\dfrac{f(3)-f(0)}{3-0}\\\\=\dfrac{3^2-0^2}{3}=\dfrac{9}{3}\\\\=3$

$\Rightarrow\ f'(c)=3\ \ \ ...(i)$

Since $f'(x)=2x$

then, at x=c, $f'(c)=2c\ \ \ ...(ii)$

From (i) and (ii), we have

$2c=3\\\\\Rightarrow\ c=\dfrac{3}{2}$

Hence, the correct option is c) $\dfrac{3}{2}$ .

4 0

2 years ago

What is the Inverse operation for sin

katen-ka-za [31]

Answer:

arcsin function

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

1/2(4-+6x)=1/3x+2/3(x+9)

KonstantinChe [14]

Answer:

The value of x for the expression is 2

Step-by-step explanation:

Given algebraic expression as :

$\frac{1}{2}$ (4 + 6 x ) = $\frac{1}{3}$ x + $\frac{2}{3}$ ( x + 9 )

Now, solving the given expression

Or, $\frac{1}{2}$ × 4 + $\frac{1}{2}$ × 6 x = $\frac{1}{3}$ x + $\frac{2}{3}$ x + $\frac{2}{3}$ × 9

Or, $\frac{4}{2}$ + $\frac{6}{2}$ × x = $\frac{1}{3}$ x + $\frac{2}{3}$ x + $\frac{18}{3}$

or, 2 + 3 × x = $\frac{x}{3}$ + $\frac{2 x}{3}$ + 6

or, 2 + 3 × x = $\frac{2 x + x}{3}$ + 6

or, 2 + 3 × x = $\frac{3 x}{3}$ + 6

Or, 2 + 3 × x = x + 6

or, 3 x - x = 6 - 2

Or, 2 x = 4

∴ x = $\frac{4}{2}$

I.e x = 2

Hence The value of x for the expression is 2 Answer

8 0

2 years ago