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4 years ago

Find the quotient of 191 divided by 5

Mathematics

2 answers:

ANEK [815]4 years ago

8 0

38.2 is the quotient!!!!
Hope this helps

bogdanovich [222]4 years ago

4 0

First off,we need to understand that the equation consist of:
Devidand(191), divisor(5), quotient and remainder, while a quotient is the answer of the equation excluding the remainder.

Therefore in this case:
191÷5
= Quotient: 38, remainder: 1

The answer is 38.

Hope it helps!

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Please help!!! I need someone to use some brain power

olya-2409 [2.1K]

Answer:

r = 4, s = 3, t = 5

Step-by-step explanation:

Solving quadratic equation:

5x² - 8x + 5 = 0
x = (-b ± √b²-4ac)/2a
x = (8 ± √8²-4*5*5)/2*5
x = (8 ± √64-100)/10
x = (8 ±√-36)/10
x = (8 ± 6i)/10
x = (4 ± 3i)/5

x = (r ± si)/t

r = 4, s = 3, t = 5

3 0

3 years ago

Set up and evaluate the optimization problems. (Enter your answers as comma-separated lists.) Find two positive integers such th

alexira [117]

Answer and Step-by-step explanation:

Let x and y be two positive integers and their sum is 14:

X + y = 14

And the sum of square of this number is:

f = x2 + y2

= x2+ (14 – x)2

Differentiate with respect to x, we get:

F’(x) = [ x2 + (14 – x)2]’ = 0

2x + 2(14-x)(-1) = 0

2x +( 28 – 2x)(-1) = 0

2x – 28 +2x = 0

2x + 2x = 28

4x = 28

X = 7

Hence, y = 14 – x = 14 -7 = 7

Now taking second derivative test:

F”(x) > 0

For x = y = 7,f reaches its maximum value:

(7)2 + (7)2 = 49 + 49

= 98

F at endpoints x Є [ 0, 14]

F(0) = 02 + (14 – 0)2

= 196

F(14) = (14)2 + (14 – 14)2

= 196

Hence the sum of squares of these numbers is minimum when x = y = 7

And maximum when numbers are 0 and 14.

6 0

3 years ago

Which graph represents the function f (x) = StartFraction 2 Over x minus 1 EndFraction + 4?

Pepsi [2]

Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

$f(x) = \dfrac{2}{x - 1} + 4$

From the given function, we have;

When x = 1, the denominator of the fraction, $\dfrac{2}{x - 1}$ , which is (x - 1) = 0, and the function becomes, $\dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty$ therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as x increases, the fraction $\dfrac{2}{x - 1}$ tends to 0, therefore as x increases, we have;

$\lim_ {x \to \infty} \dfrac{2}{(x - 1)} \to 0, and \ \dfrac{2}{(x - 1)} + 4 \to 4$

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as x becomes more negative, f(x) → 4. When x = 0, $\dfrac{2}{0 - 1} + 4 = 2$ . When x approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

5 0

3 years ago

Find the sum of -3a and 5a -3

zepelin [54]

Find the sum of -3a and 5a -3

-3a + (5a-3)

2a-3

4 0

4 years ago

Determine if the ordered pair (6, 4) is a solution to the inequality

Ugo [173]

Answer:

$\Large \boxed{\mathrm{Option \ D}}$

Step-by-step explanation:

(6, 4)

x = 6 and y = 4

y > -1/2x + 7

Plug in the values to check if it is true.

4 > -1/2(6) + 7

4 > -3 + 7

4 > 4

This statement is false.

(6, 4) lies on the line.

8 0

3 years ago