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Nina [5.8K]
3 years ago
6

I need help, im a single dad construction worker from los angeles, my daughter is a very smart person. Im having trouble correct

ing a wrong answer in math. She goes to Kumon academy for math and reading, but this problem im out of words

Mathematics
1 answer:
Deffense [45]3 years ago
6 0

Answer:

B=210

C=450

Step-by-step explanation:

Since you are asking for number 8, let's work with that problem :)

Firstly, we have to know we have rectangles here, and the area of a rectangle is given by the following formula:

A_{rectangle}=(width)(height)

So, in this problem we are given as an example the rectangle A, with witdth 90 and height 30. So, its area is:

A=(90)(30)=2700

The same happens with the small rectangle B:

B=(7)(5)=35

And, we have to do the same with rectangles B and C:

<u>Rectangle B:</u>

Width:7

Height:30

B=(7)(30)=210

<u>Rectangle C:</u>

Width:90

Height:5

B=(90)(5)=450

If we sum A, B, C and D the result must be equal to the total are of the big rectangle:

A+B+C+D=2700+210+450+35=3395

Area of big rectangle with width 90+7=97 and height 30+5=35:

(97)(35)=3395

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Answer:

cosS is 33/65

Step-by-step explanation:

7 0
3 years ago
HELP ME IT'S URGENT I COULD FAIL
hjlf

Answer:

pm+2=8

Step-by-step explanation:

Given,

m=3

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5 0
2 years ago
Read 2 more answers
(-y+5•3)+7.2y-9)=6.2y+n
kramer
(-y+5x3)+(7.2y-9)=6.2y+n
(-y+15)+(7.2y-9)=6.2y+n
since you're adding the two parentheses, you don't need to have them there
-y+15+7.2y-9=6.2y+n
7.2y-y +15-9 =6.2y+n
6.2y + 6 =6.2y+n
6.2y - 6.2y -n = -6
-n=-6
n=6

3 0
3 years ago
Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{
SVETLANKA909090 [29]

y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

  • Note 1: A more general solution is y = Ce^x for some constant C.
  • Note 2: It might be tempting to say the general solution is y = e^x+C, but that is not the case because y = e^x+C \to y' = e^x+0 = e^x and we can see that y' = y would only be true for C = 0, so that is why y = e^x+C does not work.
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3 years ago
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There will be 12 tomatoes on Thursday. 3x4
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