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9 days ago

Francine has a picture whose length is 4/3 its width. She wants to enlarge the picture to have an area of 192 in2. What will the

dimensions of the enlarged picture be? Model the scenario and solve. Then, explain in at least one sentence your solution and include the reasonableness of your solution.

Mathematics

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The area of ABED is 49 square units. Given AGequals9 units and ACequals10 units, what fraction of the area of ACIG is represent

Svet_ta [12734]

Answer:

The shaped region accounts for 7/18 of the area of ACIG.

Step-by-step explanation:

Refer to the attached diagram for further clarity on the problem.

Step 1

Determine the length of one side of square ABED.

We know that

AB=BE=ED=AD

The area of a square can be calculated as

$A=b^{2}$

where b is the side length.

We have

$A=49\ units^2$

So we substitute

$49=b^{2}$

$b=7\ units$

Thus,

$AB=BE=ED=AD=7\ units$

Step 2

Calculate the area of ACIG.

The area of rectangle ACIG is determined by

$A=(AC)(AG)$

Substituting the given values yields

$A=(9)(10)=90\ units^2$

Step 3

Determine the area of the shaded rectangle DEHG.

The area of rectangle DEHG is given by

$A=(DE)(DG)$

We find $DE=7\ units$

$DG=AG-AD=9-7=2\ units$

and substitute $A=(7)(2)=14\ units^2$

Step 4

Calculate the area of shaded rectangle BCFE.

The area of rectangle BCFE equals

$A=(EF)(CF)$

We see that

$EF=AC-AB=10-7=3\ units$

$CF=BE=7\ units$

and substitute

$A=(3)(7)=21\ units^2$

Step 5

Add the areas of the shaded regions together.

$14+21=35\ units^2$

Step 6

Divide the area of the shaded region by the area of ACIG.

$\frac{35}{90}$

Simplify this fraction by dividing both the numerator and denominator by 5.

$\frac{7}{18}$

Hence, the shaped region represents 7/18 of the area of ACIG.

5 0

2 months 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:

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

The table representing function $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ is included below.

For more information:

1. What is the y-intercept of the quadratic function f(x) = (x – 6)(x – 2)? (0,–6) (0,12) (–8,0) (2,0)

2. Which is the graph of f(x) = (x – 1)(x + 4)?

6 0

1 month ago

Solve the system. x + y = 8 2x2 – y = –5 Add the equations, then choose the result. 3x2 = –13 3x2 = 3 2x2 + x = 3 2x2 = –13

tester [12383]

2x^2 - y = -5
x + y = 8
----------------adding gives us
2x^2 + x = 3 <==

** It's important to note that 2x^2 and x cannot be combined, because they are not like terms.

6 0

2 months ago

Read 2 more answers

7. During a person's commute to school, she spends 10 minutes driving 30 miles per hour (mph) and 5 minutes stopped at red light

Inessa [12570]

Answer:

5 miles in total

Step-by-step explanation:

Given:

Time spent driving = 10 min = 10 / 60 = 1/6 hour

Duration of stop = 5 min

Driving speed = 30 miles per hour

Find:

Complete distance

Computation:

Distance traveled = Speed × time

Distance = 30 × (1/6)

Total distance = 5 miles

5 0

3 months ago