The answer to your question is y=1
Answer: The length of the rectangle is 35 in.
The width of the rectangle is 28 in.
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Explanation:
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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).
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→ 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.
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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" ;
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→ 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 ;
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Now, we can solve for the "length" and the "width" ;
→ The length is: "5x" ;
5x = 5(7) = " 35 in " ;
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→ The width is: "4x" ;
4x = 4(7) = "28 in."
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Let us check our answer:
→ A = L * w ;
→ 980 in² = ? 35 in. * 28 in. ?? ;
Using a calculator: "35 * 28 = 980" . Yes! ;
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{ Also note: " in * in = in² " ? Yes! } .
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Answer:
2√2.
Step-by-step explanation:
We use the Pythagoras Theorem:
x^2 = 2^2 + 2^2
x^2 = 8
x = √8
= 2√2.
The functions which are even are symmetric to y-axis. By even functions, it means that f(x) = f(x) for all the domain values. Thus even functions for example, x^2, 1+x^2, cosx, x^4 all follow the conditions i.e. f(x) = f(x) and thus they are symmterical to y-axis.
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -3}}\quad ,&{{ -1}})\quad % (c,d) &({{ 1}}\quad ,&{{ 2}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ r=\sqrt{[1-(-3)]^2+[2-(-1)]^2}\implies r=\sqrt{(1+3)^2+(2+1)^2} \\\\\\ r=\sqrt{16+9}\implies r=\sqrt{25}\implies r=5](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%28%7B%7B%20-3%7D%7D%5Cquad%20%2C%26%7B%7B%20-1%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%201%7D%7D%5Cquad%20%2C%26%7B%7B%202%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B%5B1-%28-3%29%5D%5E2%2B%5B2-%28-1%29%5D%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B%281%2B3%29%5E2%2B%282%2B1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B16%2B9%7D%5Cimplies%20r%3D%5Csqrt%7B25%7D%5Cimplies%20r%3D5)
![\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{\quad k}})\qquad radius=&{{ r}}\\ &-3&-1&5 \end{array} \\\\\\\ [x-(-3)]^2+[y-(-1)]^2=5^2\implies (x+3)^2+(y+1)^2=25](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%7B%7B%20h%7D%7D%29%5E2%2B%28y-%7B%7B%20k%7D%7D%29%5E2%3D%7B%7B%20r%7D%7D%5E2%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Blllll%7D%0Acenter%5C%20%28%26%7B%7B%20h%7D%7D%2C%26%7B%7B%5Cquad%20%20k%7D%7D%29%5Cqquad%20%0Aradius%3D%26%7B%7B%20r%7D%7D%5C%5C%0A%26-3%26-1%265%0A%5Cend%7Barray%7D%20%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-3%29%5D%5E2%2B%5By-%28-1%29%5D%5E2%3D5%5E2%5Cimplies%20%28x%2B3%29%5E2%2B%28y%2B1%29%5E2%3D25)