98,205 in scientific notation

((11 - 3)2 x 4) - 13 + 42<br> Help pls I will mark brainliest

tankabanditka [31]

Answer:

93

Step-by-step explanation:

3 0

3 years ago

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Charlie knows that 3 blocks are needed to make a tower that is 0.5 foot high.

iogann1982 [59]

Answer:

24 blocks will be needed to make the tower that is 4 feet high

Step-by-step explanation:

Let us use the ratio method to solve the question

∵ Charlie knows that 3 blocks are needed to make a tower that is

0.5 foot high

∵ He needs to make a tower that is 4 feet high

→ By using the ratio method

→ Blocks : height (ft)

→ 3 : 0.5

→ x : 4

→ By using the cross multiplication

∵ x × 0.5 = 3 × 4

∴ 0.5x = 12

→ Divide both sides by 0.5 to find x

∵ $\frac{0.5x}{0.5}$ = $\frac{12}{0.5}$

∴ x = 24

∵ x represents the number of blocks

∴ 24 blocks will be needed to make the tower that is 4 feet high

7 0

2 years ago

Additive inverse of 1/3

Xelga [282]

The additive inverse of 1/3 is 3/1.

Hope this helps you!

:)

4 0

3 years ago

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Sally's soccer team played 25

pantera1 [17]

Answer:

64%

Step-by-step explanation:

25 - 9 = 16, so 16 of the games were won by Sally's team.

16/25 = 64% of the games were won

6 0

2 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

2 years ago