How much time is in. 7:30 to 9:29

Mathematics

2 answers:

vitfil [10]2 years ago

5 0

It is 1 hr, 59 min from the time alotted if it is am and am or pm and pm

Nostrana [21]2 years ago

4 0

Hi

I think that the answer is 1.98 hours .

7:30 + 30 minutes = 8 o'clock
8 o'clock +60 minutes = 9 o'clock
9 o'clock + 29 minutes = 9:29

60+30= 90+29=119

119 minutes= 1.98 hours

You might be interested in

Tell me all the properties of algebra

Keith_Richards [23]

Answer:

Commutative Property of Addition.-
2 + 3 = 3 + 2

Commutative Property of Multiplication.- ex. 5 × 7 = 7 × 5

Associative Property of Addition.- ex.(2 + 3) + 6 = 2 + (3 + 6)

Associative Property of Multiplication.- ex. (7 × 3) × 10 = 7 × (3 × 10)
Distributive Properties of Addition Over Multiplication- ex. 2 × (2 + 8) = 2 × 2 + 2 × 8
(2 + 8) × 10 = 2 × 10 + 8 × 10

The reciprocal of a non zero real number a is 1/a.- ex. reciprocal of 5 is 1/5 and 5 × (1/5) = 1

Step-by-step explanation:

7 0

2 years ago

Find integra of xlnxdx

Mademuasel [1]

Answer:

$\dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

$\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\end{minipage}}$

Increase the power by 1, then divide by the new power.

Given indefinite integral:

$\displaystyle \int x \ln x \:\: \text{d}x$

To integrate the given integral, use Integration by Parts:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

$\text{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}$

$\text{Let }\dfrac{\text{d}v}{\text{d}x}=x \implies v=\dfrac{1}{2}x^2$

Therefore:

$\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x \ln x\:\:\text{d}x & = \ln x \cdot \dfrac{1}{2}x^2-\int \dfrac{1}{2}x^2 \cdot \dfrac{1}{x}\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x -\int \dfrac{1}{2}x\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}\end{aligned}$