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8 days ago

PLEASE HURRYY I AM BEGGING FOR HELP IVE BEEN STUCK FOR AN HOUR

Mathematics

1 answer:

zzz [12.3K]8 days ago

3 0

Part A:

Please refer to the attached image to view the graph illustrating the inequalities.

There are two distinct lines shown, one with a steeper negative slope than the other. The shaded region in between indicates the solution set.

The green line corresponds to 2x+y≤8, while the blue line represents x+y≥4

Part B:

To determine if (8, 10) is included, substitute x = 8 and y = 10 into both inequalities. If it fails to satisfy either inequality, then it won't be part of the solution area.

2(8) + 10 ≤ 8

16 + 10 ≤ 8

26 ≤ 8... 26 is NOT less than or equal to 8

Hence, (8, 10) cannot be a solution because the inequality does not hold.

Part C:

I will select a random point from the graph, such as (2, 3). For your answer, any point within the shaded region where both x and y are positive would suffice.

The point (2, 3) indicates that Sarah can purchase 2 cupcakes and 3 pieces of fudge, ensuring that she acquires at least 4 pastries for her siblings while adhering to her 8 dollar budget.

Please let me know if you need any further clarification. Happy studying!

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28 days ago

Two functions are shown in the table below. Function 1 2 3 4 5 6 f(x) = −x2 + 4x + 12 g(x) = −x + 6 Complete the table on your o

Svet_ta [12734]

For $\fbox{\begin \\\math{x}=6\\\end{minispace}}$ the function $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ both yield the same result.

Detailed breakdown:

The functions involved are

$f(x)=-x^{2}+4x+12$

$g(x)=-x+6$

Step 1:

Insert $x=1$ in $f(x)=-x^{2} +4x+12$ to find the value of $f(1)$ .

$f(1)=-1^{2} +4(1)+12\\f(1)=-1+4+12\\f(1)=15$

Insert $x=1$ in $g(x)=-x+6$ to find the value of $g(1)$ .

$g(1)=-1+6\\g(1)=5$

Step 2:

Insert $x=2$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(2)$ .

$f(2)=-2^{2} +4(2)+12\\f(2)=-4+8+12\\f(2)=16$

Substitute $x=2$ into $g(x)=-x+6$ to find the value of $g(2)$ .

$g(2)=-2+6\\g(2)=4$

Step 3:

Replace $x=3$ in $f(x)=-x^{2} +4x+12$ to find the value of $f(3)$ .

$f(3)=-3^{2} +4(3)+12\\f(3)=-9+12+12\\f(3)=15$

Also, replace $x=3$ in $g(x)=-x+6$ to find the value of $g(3)$ .

$g(3)=-3+6\\g(3)=3$

Step 4:

Insert $x=4$ in $f(x)=-x^{2} +4x+12$ to find the value of $f(4)$ .

$f(4)=-4^{2} +4(4)+12\\f(4)=-16+16+12\\f(4)=12$

Also, replace $x=4$ in $g(x)=-x+6$ to obtain the value of $g(4)$ .

$g(4)=-4+6\\g(4)=2$

Step 5:

Insert $x=5$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(5)$ .

$f(5)=-5^{2} +4(5)+12\\f(5)=-25+20+12\\f(5)=7$

Replace $x=5$ in $g(x)=-x+6$ to find the value of $g(5)$ .

$g(5)=-5+6\\g(5)=1$

Step 6:

Insert $x=6$ into $f(x)=-x^{2} +4x+12$ to find the value of $f(6)$ .

$f(6)=-6^{2} +4(6)+12\\f(6)=-36+24+12\\f(6)=0$

Also, substitute $x=6$ in $g(x)=-x+6$ to obtain the value of $g(6)$ .

$g(6)=-6+6\\g(6)=0$

Step 7:

According to the provided condition $f(x)=g(x)$ .

(a). Insert $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ into the previously mentioned equation.

$-x^{2} +4x+12=-x+6$

(b). Multiply through by $-1$ on both sides.

$x^{2} -4x-12=x-6$

(c). Move the term $x-6$ to the left side of the equation.

$x^{2} -4x-12-x+6=0\\x^{2} -5x-6=0$

(d). Divide the middle term so that its sum equals 5 and the product equals 6.

$x^{2} -(6-1)x-6=0\\x^{2} -6x+x-6=0\\x(x-6)+1(x-6)=0\\(x+1)(x-6)=0\\x=-1,6$

From the analysis above, it is noted that for $x=6$ both functions $f(x)$ and $g(x)$ yield the same outcome.

Using a direct approach:

$f(x)=g(x)\\\Leftrightarrow-x^{2} +4x+12=-x+6\\\Leftrightarrow-x^{2} +4x+12+x-6=0\\\Leftrightarrow-x^{2} +5x+6=0\\\Leftrightarrow-x^{2} +6x-x+6=0\\\Leftrightarrow x^{2} -6x+x-6=0\\\Leftrightarrow x(x-6)+1(x-6)=0\\\Leftrightarrow(x+1)(x-6)=0\\\Leftrightarrow x=6,-1$