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3 years ago

Find the distance beyween S(6, -3) and T (7, -2)

Mathematics

2 answers:

Tasya [4]3 years ago

7 0

I believe that the answer is 2

s344n2d4d5 [400]3 years ago

3 0

Answer:

$d=\sqrt{2}$

Step-by-step explanation:

Distance Formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Simply plug in your 2 coordinates into the distance formula to find distance <em>d</em>: $d=\sqrt{(7-6)^2+(-2-(-3))^2}$

$d=\sqrt{(1)^2+(-2+3)^2}$

$d=\sqrt{1+(1)^2}$

$d=\sqrt{1+1}$

$d=\sqrt{2}$

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Answer:

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3 years ago

In ΔSTU, the measure of ∠U=90°, ST = 9.7 feet, and US = 4 feet. Find the measure of ∠T to the nearest degree.

BaLLatris [955]

Answer:

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*Make sure calculator is on degree mode*

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3 years ago

Find the EXACT value of sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13 where B is in Quadrant IV. Assume al

MissTica

$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)$

well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.

$\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5$

$\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}$

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3 years ago

Solve 2x2 − 8x = −7.

leva [86]

$\sf 2x^2 - 8x = - 7 \\ \\ Subtract \ -7 \ from \ both \ sides \ of \ the \ equation. \\ \\ 2x^2 - 8x - (-7 ) = - 7 - (-7) \\ \\ 2x^2 - 8x + 7 = 0 \\ \\ Use \ quadratic \ formula \ a = 2, b = -8, c = 7 \\ \\ x = \dfrac{- b \pm \sqrt{b^2- 4ac} }{2a} \\ \\ \\ x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 (2) (7) } }{2(2)} \\ \\ x = \dfrac{8 \pm \sqrt{8} }{4} \\ \\ x = 2 + \frac{1}{2}\sqrt{2} \ or \ x = 2 + \frac{-1}{2} \sqrt{2}$

Ur answer is the fourth one

7 0

4 years ago

Read 2 more answers