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bazaltina [42]
3 years ago
5

A smaller square is located inside a larger square the length of the smaller square is seven CM and length of the larger square

is 12 CM find the area of the selection outside the small square but inside the larger square PLZ HELLPPP

Mathematics
2 answers:
ohaa [14]3 years ago
8 0

Answer: The answer would be the area of the large square subtracted by the small square.

12 x 12 = 144

7 x 7 = 49

144 - 49 = 95

The answer is 95.

Maurinko [17]3 years ago
5 0

find the area of the little square then find the area of the big square and sub tract the little square area from the big square

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The slope of XY is the same as the slope of AB because AB is parallel to XY and parallel lines have the same slope.
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10. What is the value of ? A. 47.5° B, 85° 95° C. 95° С D. 132.59​
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What is the area of a triangle whose vertices are X(1, 1), Y(3, -1), and Z(4,4)?
shepuryov [24]

Answer:

Step-by-step explanation:

Note that in this diagram, point A represents point X, point B represents point Y, and point C represents point Z.

From the diagram, it appears as if \overline{XZ} \perp \overline{XY}. To determine if this is the case, we can find the slopes of both segments.

m_{\overline{XZ}}=\frac{4-1}{4-1}=1\\m_{\overline{XY}}=\frac{-1-1}{3-1}=-1

Since these slopes are negative reciprocals, we know that they are perpendicular.

This means we can use the formula A=\frac{1}{2}bh, where b is the base (XY) and h is the height (XZ).

  • Note these are interchangeable.

Using the distance formula,

XY=\sqrt{(3-1)^{2}+(-1-1)^{2}}=2\sqrt{2}\\XZ=\sqrt{(4-1)^{2}+(4-1)^{2}}=3\sqrt{2}

This means the area is \frac{1}{2}(2\sqrt{2})(3\sqrt{2})=\boxed{6}

7 0
2 years ago
What is the solution of -8/2y-8=5/y+4 - 7y+8/y^2-16
blsea [12.9K]

Answer:

<h2>y = 8</h2>

Step-by-step explanation:

Domain:\\\\2y-8\neq0\ \wedge\ y+4\neq0\ \wedge\ y^2-16\neq0\\\\2y\neq8\ \wedge\ y\neq-4\ \wedge\ y^2\neq16\\\\y\neq4\ \wedge\ y\neq-4\ \wedge\ y\neq\pm\sqrt{16}\\\\y\neq4\ \wedge\ y\neq-4\ \wedge\ y\neq-4\ \wedge\ y\neq4\\\\\boxed{y\neq-4\ \wedge\ y\neq4}\\\\===========================

\dfrac{-8}{2y-8}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-16}\\\\\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-4^2}\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{(y-4)(y+4)}\qquad\text{multiply both sides by (-2)}\\\\\dfrac{8}{y-4}=-\dfrac{10}{y+4}+\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{add}\ \dfrac{10}{y+4}\ \text{to both sides}\\\\\dfrac{8}{y-4}+\dfrac{10}{y+4}=\dfrac{14y+16}{(y-4)(y+4)}

\dfrac{8(y+4)}{(y-4)(y+4)}+\dfrac{10(y-4)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\\\\\dfrac{8(y+4)+10(y-4)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{use the distributive property}\\\\\dfrac{8y+32+10y-40}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{combine like terms}\\\\\dfrac{(8y+10y)+(32-40)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\\\\\dfrac{18y-8}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\iff18y-8=14y+16\\\\18y-8=14y+16\qquad\text{subtract 14y from both sides}

4y-8=16\qquad\text{add 8 to both sides}\\\\4y=24\qquad\text{divide both sides by 4}\\\\y=8\in D

8 0
3 years ago
What is the solution for this equation -9y=9
Valentin [98]
Divide both sides of this eq'n by -9.  You'll get y = -1.

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