All categories

3 years ago

Me ayudan a resolver unos problemas en word les adjunto el archivo

1 answer:

3 0

Hola! yo te ayudaria, pero no se en que. No tienes el problema para que yo lo resuelva. Lo siento si mi espanol no es bueno, estoy aprendiendo como escribir lo. Sorry-

You might be interested in

Write an expression to represent the sum of three times the square of a number and -7 . In your expression what is the value of

stich3 [128]

Answer:

Step-by-step explanation:

3x^2-7

8 0

3 years ago

The cost of a jacket was originally 40$ It was on sale for 20% how much did they save

ASHA 777 [7]

Answer:

Step-by-step explanation:

In this circumstance, 20% off is the original number (40) divided by 5, and that's the answer. I hope this helps!

7 0

3 years ago

Read 2 more answers

There are x number of students at helms. If the number of students increases by 7.8% each year, how many students will be there

vodomira [7]

There will be 1.078x students next year and equation is number of students in next year = x + 7.8% of x

<h3><u>Solution:</u></h3>

Given, There are "x" number of students at helms.

The number of students increases by 7.8% each year which means if there "x" number of students in present year, then the number of students in next year will be x + 7.8% of x

Number of students in next year = number of students in present year + increased number of students.

$\begin{array}{l}{\text { Number of students in next year }=x+7.8 \% \text { of } x} \\\\ {\text { Number of students in next year }=x\left(1+\frac{7.8}{100}\right)} \\\\ {\text { Number of students in next year }=x(1+0.078)=1.078 x}\end{array}$

Thus there will be 1.078x students in next year

3 0

3 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

adoni [48]

Answer: The required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Step-by-step explanation:

Since we have given that

$y=\ln[x(2x+3)^2]$

Differentiating log function w.r.t. x, we get that

$\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Hence, the required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

3 0

3 years ago

4 Compare 5.269 and 5.038.

Hunter-Best [27]

Answer:

the answer is B

Step-by-step explanation:

4 0

3 years ago