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FromTheMoon [43]
3 years ago
14

Which equation represents a line which is perpendicular to the line 6x+y=1

Mathematics
1 answer:
Andre45 [30]3 years ago
6 0
I believe it’s A because if it is perpendicular you would have to flip 6X and change the sign so therefore it would be A. In slope intercept form it would be y=-6x+1.
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Help please! im not sure which graph is which
Wewaii [24]

WHat do you mean when you say you don't know "which graph is which"

8 0
3 years ago
If J/24 = K, then J/6 =
tensa zangetsu [6.8K]

Answer:

4K

Step-by-step explanation:

\frac{J}{24} = K ( multiply both sides by 24 to clear the fraction )

J = 24K ( divide both sides by 6 )

\frac{J}{6} = 4K

8 0
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}

4 0
2 years ago
For what values of x and y must ABCD be a parallelogram?
n200080 [17]
<h2>x^{\circ}=21^{\circ} \ and \ y^{\circ}=39^{\circ}</h2>

Step-by-step explanation:

We know that vertically opposite angles of a parallelogram are equal .

\angle ABC =\angle ADC\ and \ \angle BAD = \angle DAC

⇒3x^{\circ}=(4x-21)^{\circ} \ and \ 3y^{\circ}=(y+78)^{\circ}

⇒3x^{\circ}-4x^{\circ}=-21^{\circ} \ and \ 3y^{\circ}-y^{\circ}=78^{\circ}

⇒-x^{\circ}=-21^{\circ} \ and \ 2y^{\circ}=78^{\circ}

⇒x^{\circ}=21^{\circ} \ and \ y^{\circ}=39^{\circ}

7 0
3 years ago
Casey is completing an experiment that requires 300 mL of potassium chloride (KCl) at a concentration of 0.2 M. But Casey's lab
gulaghasi [49]
What you first need to do is make an equation. Then plug that equation into photomath and you get your answer


3 0
3 years ago
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