Take the area of the small photograph and divide it by four, assuming it's a square, each side would then be 6 inches. The large photograph has double the side length of the small one, so 6 multiplied by 2 is 12. To find the area of the large (square) photograph, multiply 12 by 12 as two of the sides, and the area of the large photo is 144 square inches
Answer:
where is her work?
Step-by-step explanation:
maybe you forgot to put the pic
Answer: The length of the rectangle is 35 in.
The width of the rectangle is 28 in.
_________________________________________________________
Explanation:
_________________________________________________________
The formula for the area, "A", of a rectangle:
Area (A) = length (L) * width (w) ;
that is: " A = L * w " ;
A = 980 in² (given);
ratio of the length to the width is: " 5 : 4 " (given);
→ Find the length (L) and the width (w).
_________________________________________________________
→ 980 in² = (5x) * (4x) ;
in which: " 980 in² " is the area of the triangle;
" 5x" = the length (L) of the rectangle, for which we shall solve;
" 4x" = the width (w) of the rectangle, for which we shall solve.
_______________________________________________________
If we find solve for "x" ; we can solve for "5x" and "4x" (the "length" and the "width", respectively); by plugging in the solved value for "x" ;
_______________________________________________________
→ 980 in² = (5x) * (4x) ;
↔ (5x) * (4x) = 980 in² ;
→ (5x) * (4x) = (5) * (4) * (x) * (x) = 20 * x² = 20x² ;
→ 20x² = 980 ;
Divide each side by "10" ; by canceling out a "0" on each side of the equation:
→ 2x² = 98 ;
Now, divide each side of the equation by "2" ;
→ 2x² / 2 = 98 / 2 ;
to get:
→ x² = 49 ;
Now, take the "positive square root" of each side of the equation; (since a "length or width" cannot be a "negative value") ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ ⁺√(x²) = √49 ;
to get:
→ x = 7 ;
______________________________________________________
Now, we can solve for the "length" and the "width" ;
→ The length is: "5x" ;
5x = 5(7) = " 35 in " ;
______________________________________________________
→ The width is: "4x" ;
4x = 4(7) = "28 in."
______________________________________________________
Let us check our answer:
→ A = L * w ;
→ 980 in² = ? 35 in. * 28 in. ?? ;
Using a calculator: "35 * 28 = 980" . Yes! ;
____________________________________________________
{ Also note: " in * in = in² " ? Yes! } .
____________________________________________________
H(t) = -16t² + 60t + 95
g(t) = 20 + 38.7t
h(1) = -16(1²) + 60(1) + 95 = -16 + 60 + 95 = -16 + 155 = 139
h(2) = -16(2²) + 60(2) + 95 = -16(4) + 120 + 95 = -64 + 215 = 151
h(3) = -16(3²) + 60(3) + 95 = -16(9) + 180 + 95 = -144 + 275 = 131
h(4) = -16(4²) + 60(4) + 95 = -16(16) + 240 + 95 = -256 + 335 = 79
g(1) = 20 + 38.7(1) = 20 + 38.7 = 58.7
g(2) = 20 + 38.7(2) = 20 + 77.4 = 97.4
g(3) = 20 + 38.7(3) = 20 + 116.1 = 136.1
g(4) = 20 + 38.7(4) = 20 + 154.8 = 174.8
Between 2 and 3 seconds.
The range of the 1st object is 151 to 131.
The range of the 2nd object is 97.4 to 136.1
h(t) = g(t) ⇒ 131 = 131
It means that the point where the 2 objects are equal is the point where the 1st object is falling down while the 2nd object is still going up.
