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3 years ago

Perform the indicated operation. 61/2 · 4/5? a.5 1/5 b.6 2/5 c.3 1/5 d.4 2/5

Mathematics

2 answers:

dexar [7]3 years ago

5 0

Answer:

5 1/5

Step-by-step explanation:

6 1/2 *4/5

Rewriting as an improper fraction

( 2*6 +1)/2 *4/5

13/2 *4/5

52/ 10

50/10 + 2/10

5 +1/5

5 1/5

Bas_tet [7]3 years ago

4 0

Answer:

A. 5 1/5

Step-by-step explanation:

6 1/2 × 4/5

Convert 6 1/2 to an improper fraction.

13/2 × 4/5

Multiply both fractions.

52/10

Convert the improper fraction to a mixed fraction.

5 2/10

⇒ 5 1/5

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Amber saved a total of $3.20 over 5 weeks. She saved the same amount of money each week.

dybincka [34]

Answer:

0.64

Step-by-step explanation:

you just divide 3.20 by 5

3 0

3 years ago

Read 2 more answers

I need help on this

IRINA_888 [86]

Answer:

Rate of change is slope

To calculate slope its

y2 - y1 / x2- x1 = slope

Just choose 2 random sets from the table.

I’m gonna go with (0,-10) and (2,20)

Now calculate slope

20 - (-10) / 2 - 0

Solve.

30 / 2

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7 0

2 years ago

Hello, I'm trying to solve 6x + y = 4, x - 4y = 19. I have found the y but I cant seem to find the x. Plz help and we cant use d

gizmo_the_mogwai [7]

Answer:

$x=\frac{7}{5}$

$y=-\frac{22}{5}$

Step-by-step explanation:

Given:

The given expressions are.

$6x+y=4$

$x-4y=19$

We need to find x and y values.

Solution:

Equation 1⇒ $6x+y=4$

Equation 2⇒ $x-4y=19$

First solve the equation 1 for y.

$6x+y=4$

$y = 4-6x$ --------(3)

Substitute $y = 4-6x$ in equation 2.

$x-4(4-6x)=19$

Simplify.

$x-(4\times 4 - 4\times 6x)=19$

$x-(16-24x)=19$

$x-16+24x=19$

Add 16 both side of the equation.

$25x-16+16=19+16$

$25x=35$

$x=\frac{35}{25}$

Divide Numerator and denominator by 5.

$x=\frac{7}{5}$

Substitute x value in equation 3 and simplify.

$y=4-6(\frac{7}{5})$

$y=4-\frac{6\times 7}{5}$

$y=4-\frac{42}{5}$

$y=\frac{5\times 4-42}{5}$

$y=\frac{20-42}{5}$

$y=-\frac{22}{5}$ .

Therefore, the value of $x=\frac{7}{5}$ and $y=-\frac{22}{5}$ .

6 0

4 years ago

Please help me with 2b ASAP. <br> Really appreciate it!!

Bogdan [553]

$f(x)=\dfrac{x^2}{x^2+k^2}$

By definition of the derivative,

$f'(x)=\displaystyle\lim_{h\to0}\frac{\frac{(x+h)^2}{(x+h)^2+k^2}-\frac{x^2}{x^2+k^2}}h$

$f'(x)=\displaystyle\lim_{h\to0}\frac{(x+h)^2(x^2+k^2)-x^2((x+h)^2+k^2)}{h(x^2+k^2)((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{(x+h)^2-x^2}{h((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{2xh+h^2}{h((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{2x+h}{(x+h)^2+k^2}$

$f'(x)=\dfrac{2xk^2}{(x^2+k^2)^2}$

$\dfrac{k^2}{(x^2+k^2)^2}$ is positive for all values of $x$ and $k$ . As pointed out, $x\ge0$ , so $f'(x)\ge0$ for all $x\ge0$ . This means the proportion of occupied binding sites is an increasing function of the concentration of oxygen, meaning the presence of more oxygen is consistent with greater availability of binding sites. (The question says as much in the second sentence.)