All categories

3 years ago

How does knowing the double 6+6=12 help you solve the near double 6+7=13

Mathematics

1 answer:

forsale [732]3 years ago

8 0

Answer:

6 + 7 = 13

Step-by-step explanation:

We are given the following information in the question:

We know that

$6 + 6 = 12$

We need this information to evaluate

$6+7$

The evaluation can be done with the help of using associative property and distributive property.

Associative property: (a + b) + c = a + (b + c)

The evaluation can be shown as:

$6 + 7\\= 6 + 6 + 1\\=(6 + 6) + 1\\= 12 + 1\\=13$

You might be interested in

Marcus was training for a 2,000 meter race. He had to alternate between running and walking.

evablogger [386]

Answer:

if marcus ran 1,600 meters that means he walked 400 meters

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Please help, I’ll mark as brainliest!! The picture is above :)

Ray Of Light [21]

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable

Exact Form:

x = - 25/8 $\frac{x}{y}$

Decimal Form:

x = -3.125

Mixed Number Form:

x = - 3/1/8 $\frac{x}{y}$

Step-by-step explanation:

5 0

3 years ago

The Kansas predicted that in 2018 the population in Centerville would be 17,466 and actuality it was 16,400 what is the percent

sammy [17]

Answer:

6.5%

Step-by-step explanation:

Percentage error formula =

|Estimated value - Real value|/Real value × 100

From the above question,

Estimated value = 17,466

Real value = 16,400

Percentage error = |17,466 - 16,400|/16,400 × 100

= 1066/16,400 × 100

= 0.065 × 100

= 6.5%

Therefore, the percent error = 6.5%

5 0

3 years ago

Read 2 more answers

2. On a number line, G = 8 and H = –3. If H is the midpoint<br><br>of GI , find the coordinate of I.

MissTica

Answer:

-14

Step-by-step explanation:

the coordinate of I :

H = 8+I /2

-3 = 8+I /2

8+I = 2×(-3)

8+ I = -6

I = -6-8

I = -14

8 0

3 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

3 years ago