All categories

3 years ago

By how much is 3 4/5 greater than the product of 3/5 x 1 3/4

Mathematics

1 answer:

dalvyx [7]3 years ago

7 0

Answer: 6.65

Step-by-step explanation:

3 4/5 = 3.8

1 3/4 = 1.75

3.8*1.75= 6.65

Hope it helps <3

You might be interested in

Plz help me with dis its one question

tatiyna

Answer:

Alix does. See below.

Step-by-step explanation:

Givens

Jenny

b1 = 90 cm
b2 = 80 cm
height = 23 cm

Alex

b1 = 90 cm
b2 = 70 cm
height = 27 cm

Formula

For both territories the answer is based on the formula

Area = (b1 + b2)*h/2

Solution

Jenny

Area = (90 + 80)23/2
Area = 170 * 23 / 2
Area = 85 * 23
Area = 1955 units.

Alex

Area = (70 + 90)*27/2
Area = 160*27 / 2
Area = 80* 27
Area = 2160

Alix does by 205 cm

7 0

3 years ago

I kinda forgot how to do things like 2 divided by 5. plz help

pentagon [3]

Answer:

the answer 0.4

Step-by-step explanation;

4 0

3 years ago

Find the x-intercept and y-intercept of the line. 3x + y = 6

Dafna1 [17]

Answer:

x = (6 - y)÷ 3

Step-by-step explanation:

3x + y = 6

3x = 6 - y

x = 6 - y ÷ 3

6 0

3 years ago

Question 4 pleaseeeeeee

arsen [322]

Answer:

Step-by-step explanation:

Unit rate means one quantity in terms of another unit quantity. In the answer choices, we see that the denominator is always in terms of UNIT LITERS (1 L), so that means, we have to find gallons in terms of UNIT LITERS.

It means, we have to divide gallons by liters to get in terms of unitary liters.

Let's check the first line:

19.5/75 = 0.26 Gallons / 1 Liter

Now, checking 2nd:

39/150 = 0.26 Gallons / 1 Liter

So, yes, they are the same and other 2 lines would be the same as well. The unit rate is:

0.26 gal/1L

That's answer choice B

7 0

3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers