The 4•x by 5•x inner dimensions of the 53 inches outer perimeter picture frame, the 1 inch wide frame and the 4 × 5 inch picture to be enlarged, gives;

a. The outer perimeter equation is presented as follows;

18•x + 8 = 53 inches

b. The scale factor required for the enlargement of the picture is 2.5

<h3>Which method can be used to write the equation and find the scale factor?</h3>

The dimensions of the photo = 4 × 5 inch

Width of the frame = 1 inch

Outer perimeter of the frame = 53 inches

The width of the inside of the picture frame = 4•x

The height of the inside of the picture frame = 5•x

a. The equation for the outer perimeter of the picture frame is therefore;

2 × (4•x + 1 + 1) + 2 × (5•x + 1 + 1) = 53

Which gives;

2 × (4•x + 2) + 2 × (5•x + 2) = 53

8•x + 4 + 10•x + 4 = 53

A simplified equation for the outer perimeter of the picture frame is therefore;

18•x + 8 = 53

b. Solving the above equation, we have;

18•x + 8 = 53

18•x = 53 - 8 = 45

18•x = 45

Therefore;

x = 45 ÷ 18 = 2.5

x = 2.5

The width and height of the picture frame are therefore;

Width = 4•x = 4 × 2.5 = 10

Height = 5•x = 5 × 2.5 = 12.5

Which gives;

Width = 10 inches

Height = 12.5 inches

The enlargement factor, sf, is given by the ratio of a side of the picture frame to a side of the picture as follows;

sf = 12.5/5 = 2.5

The factor by which to enlarge the picture frame is 2.5

Learn more about linear scale factors here:

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3 0

1 year ago

The water in a river moves south at 10 mi/hr. A motorboat travels due east at a speed of 20 mi/hr relative to the shore. Determi

weeeeeb [17]

Answer:22.36 mi/hr

Step-by-step explanation:

Given

water in a river moves south at $(V_w)10 mi/hr$

motorboat travels due to east at a speed of $(V_b)20 mi/hr$

Velocity of boat relative to the moving water $V_{bw}$

$V_{bw}=V_b-V_w$

$V_{bw}=20i-(-10j)=20i+10j$

Magnitude of $V_{bw}=\sqrt{20^2+10^2}=\sqrt{500}$

$V_{bw}=10\sqrt{5} \approx 22.36 mi/hr$

$tan\theta =\frac{10}{20}$

$\theta =26.56^{\circ}$ with the x axis.

4 0

2 years ago

Find the equation of a circle with a point at ( 10 , - 4 ) and a point at ( -2 , - 4 )

astra-53 [7]

Answer:

Solution given:

letA=(10,-4)

B=(-2,-4)

centre[C](h,k)= $\frac{10-2}{2},\frac{-4-4}{2}=(+4,-4)$

radius= $\sqrt{(4-10)²+(-4+4)²}=6$ units

we have

Equation of a circle is;

(x-h)²+(y-k)²=r²

(x-4)²+(y+4)²=36

or.

x²-8x+16+y²+8y+16=36

x²-8x+8y+y²=36-32

x²-8x+8y+y²=4

The equation is (x-4)²+(y+4)²=36 or x²-8x+8y+y²=4.

8 0

2 years ago

Read 2 more answers

Please help me please

Ne4ueva [31]

Answer:

104 + 3(90.50)

Step-by-step explanation:

Always do what's inside the parenthesis first.

PEMDAS

4 0

3 years ago