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loris [4]
4 years ago
12

The sum of two numbers is no more than 28. Let x represent the first number, and let y represent the second number. Which

Mathematics
1 answer:
weeeeeb [17]4 years ago
6 0

Answer:

C

Step-by-step explanation:

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Resolva os sistemas abaixo <br> 2x+3y=80<br> 3x+2y=70<br><br><br><br> X=8y<br> X+4y=480
timurjin [86]
Solución 1: 2(x) + 3(y) = 80
solución 2: 3(x) + 2(y) = 70

3(y) + 2(x) = 80
2(y) + 3(x) = 70

2y = -3x + 70
y = -3x÷2 + 35

2x + 3(-3x÷ 2+35) = 80  
-5x÷2 = -25  
-5x = -50

5x = 50   
x-intercept: 10 
x = 10  
y = -3x÷ 2+35

  y = -(3÷2)(10)+35 = 20 

x-interpcet: 10
y-intercept: 20
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
solución: x - 8y = 0
solución x + 4y = 480

-8y + x = 0
4y + x = 480

x-intercept (solución 2)
x = -4x + 480

x-intercept (solución 1)
4y + 180 - 8y = 0
-12y = 480

y-intercept: 40
x-intercept: 320


Buena Suerte!


7 0
3 years ago
The value of the definite integral / 212x – sin(x) dx / 122x-sin(x) is
blondinia [14]

Hi there!

\boxed{= 70 + cos(12) - cos(2) \approx 71.26}

\int\limits^{12}_{2} {x-sin(x)} \, dx

We can evaluate using the power rule and trig rules:

\int x^n = \frac{x^{n+1}}{n+1}

\int x = \frac{1}{2}x^{2}

\int -sin(x) = cos(x)

Therefore:

\int\limits^{12}_{2} {x-sin(x)} \, dx = [\frac{1}{2}x^{2}+cos(x)]_{2}^{12}

Evaluate:

(\frac{1}{2}(12^{2})+cos(12)) - (\frac{1}{2}(2^2)+cos(2))\\= (72 + cos(12)) - (2 + cos(2))\\\\= 70 + cos(12) - cos(2) \approx 71.26

3 0
3 years ago
A right rectangular prism has a length of 7 centimeters, a width of 5 centimeters, and a height of 4 centimeters.
UkoKoshka [18]

v = b x w x h

7 x 5 x 4 = 140 cm cubed

6 0
3 years ago
Read 2 more answers
With the introduction of decimals, Max is very confused about the role zeros play in a number. For example 3 and 30 are not equa
olga2289 [7]
.3 is represented as a tenth it’s 3 out of ten, but .03 is a hundredth representing as a 3 out of a hundred it’s a relatively smaller number.
7 0
3 years ago
The triangles are similar. Write a similarity statement for the triangles.
antoniya [11.8K]

Hope this helps you

.......

4 0
2 years ago
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