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2 years ago

5/8 divided by 2 I need this fast!

Mathematics

2 answers:

Annette [7]2 years ago

6 0

Answer:

1 1/4

Step-by-step explanation:

5/8 ÷ 2

Copy dot flip

5/8 * 2/1

Rewriting

5/1 * 2/8

5 * 1/4

5/4

1 1/4

White raven [17]2 years ago

4 0

$\large \color{lime}Answer:$

$\frac{5}{8} \div 2 = \frac{5}{8} \times \frac{1}{2} = \frac{5}{16}$

$\large \color{gold}Explaination:$

Change divide into multiplication then find the reciprocal of 2 then change the places of it then multiply to 5/8 and you will get the product.

You might be interested in

complete the solution of the equation. Find the value of y when x equals 9. 3x + 4y = 43 Enter the correct answer.

3241004551 [841]

3(9)+4y=43
27+4y=43
4y=16
Y=4

4 0

3 years ago

Parallel / Perpendicular Practice

deff fn [24]

The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

Neither
║
Neither
⊥
║
Neither
Neither
Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; $m_1 = -\dfrac{1}{m_2}$

1. The given equations are in the slope and intercept form

$\ y = 3 \cdot x + 1$

The slope, m₁ = 3

$y = \dfrac{1}{3} \cdot x + 1$

The slope, m₂ = $\dfrac{1}{3}$

Therefore, the equations are neither parallel or perpendicular

Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

$y = 5 \cdot x -\dfrac{7}{2}$

Therefore, the two equations are parallel

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

$y = 2 -\dfrac{1}{2} \cdot x$

Slope, m₁ = $-\dfrac{1}{2}$

$y = \dfrac{1}{2} \cdot x - 2$

Slope, m₂ = $\dfrac{1}{2}$

The slopes

Therefore, m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

The lines are Neither parallel nor perpendicular

Neither

4. The given equations are;

2·y - x = 2

$y = \dfrac{1}{2} \cdot x +1$

m₁ = $\dfrac{1}{2}$

y = -2·x + 4

m₂ = -2

Therefore;

$m_1 \neq -\dfrac{1}{m_2}$

Therefore, the lines are perpendicular

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, $y = \dfrac{3}{4} \cdot x + 3$

Second equation, $y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}$

Therefore, m₁ = m₂, the lines are parallel

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = $\dfrac{13}{5}$

Therefore, the two equations are neither parallel nor perpendicular

Neither

7. The equations are;

2·x + 6·y = -3

Which gives $y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}$

12·y = 4·x + 20

Which gives

$y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}$

m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

Neither

8. 2·x - 5·y = -3

Which gives; $y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}$

5·x + 27 = 6

$x = -\dfrac{21}{5}$

Therefore, the slopes are not equal, or perpendicular, the correct option is Neither

Learn more here:

brainly.com/question/16732089

6 0

3 years ago

The sum of two consecutive natural numbers is 525. Find the numbers.

Luda [366]

Answer:

262 and 263 are consecutive natural numbers whose sum is 525.

Step-by-step explanation:

Natural number are a subset of real numbers consisting in positive integers (i.e. 1, 2, 3, 525) and two natural numbers are consecutive when distance between each other is equal to 1. ( $|x-y| = 1$ , $\forall\,x,y \in \mathbb{N}$ ) We can interpret the statement mathematically by means of the following expression:

$x + y = 525$

$x = n$

$y = n+1$

Where $x$ , $y$ , $n$ $\in \mathbb{N}$ .

This system is reduced to a single equation and solved afterwards:

$n +(n+1) = 525$

$2\cdot n + 1 = 525$

$2\cdot n = 524$

$n = 262$

Now we evaluate $x$ and $y$ to obtain the natural numbers:

$x = 262$

$y = 263$

262 and 263 are consecutive natural numbers whose sum is 525.

3 0

3 years ago

I think I know this much so far for A (but I could be wrong):

Eduardwww [97]

Part A. You have the correct first and second derivative.

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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

-------------------------------------------------------------

Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0.
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well.

6 0

3 years ago

If f(x) = 3x2 + 1 and g(x) = 1 – x, what is the value of (f – g)(2)?

Stels [109]

So (f-g)(x) = 3x2 + x, so when x = 2, the function is 14

4 0

3 years ago