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Arte-miy333 [17]
2 years ago
6

On the grid draw the graph of x+2y=7 for values of x between -2 and 3

Mathematics
1 answer:
lora16 [44]2 years ago
6 0

Answer: look at the picture

Step-by-step explanation: Hope this help :D

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4. The diagram at right shows the graph of 3x + 4y = 12. The shaded figure is a square, three of whose vertices are on the coord
nikklg [1K]

Remark

This is a very interesting question.  Draw a line from the origin to where the upper right vertex of the square touches the line. That line has the property that the its equation is y = x. So the "solution" to the point of intersection is the solution of the two equations.

y =  x            (1)

3x + 4y = 12 (2)

Put x in for y in equation 2

3x + 4x = 12

7x = 12

x = 12/7

x = 1.714

y = 1.714

Problem A

<em><u>x intercept</u></em>

The x intercept occurs when y = 0

3x + 4(0) = 12

3x = 12                Divide by 3

x = 12/3

x = 4

the x intercept = (4,0)

<em><u>y intercept</u></em>

The y intercept occurs when x =0

3(0) + 4y = 12

4y = 12

y = 12/4

y = 3

y intercept = (0,3)

Problem B

x and y both equal 1.714 so they are also the length of the square's side.

Problem C

See solution above. x =y is the key fact.

x = y = 1.714

5 0
3 years ago
A tree factor for 451
AlexFokin [52]
11, 41
Hope this helped 
8 0
3 years ago
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Two students Stella and vladmir complete the conversion statement 12 feet 8 inches = blank inches.
yuradex [85]

Answer:

so stella is right if thats what you ae asking

Step-by-step explanation:

12 times 12 is 144 and plus 8 is 152


3 0
3 years ago
Prove 2√(x) + 1/√(x + 1) &lt;= 2√(x+1) for all x in [0,inf)
EleoNora [17]
Start by multiplying each side of the inequality by \sqrt{x + 1} and simplifying:

2 \sqrt x + \frac{1}{\sqrt{x+1}}  \leq  2 \sqrt{x + 1}
(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})
2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)
2 \sqrt{x^2 + x} + 1 \leq 2x + 2
2 \sqrt{x^2 + x} \leq 2x + 1
\sqrt{x^2 + x} \leq x + \frac{1}{2}

From here, we can square both sides to get

x^2 + x \leq (x + \frac{1}{2})^2
x^2 + x \leq x^2 + x + \frac{1}{4}
0  \leq \frac{1}{4}, which is always true.
3 0
3 years ago
Solve the equation 1.2x = 5.16 for x.
Kobotan [32]

1.2x = 5.16

x = 5.16/1.2

x = 4.3

5 0
3 years ago
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