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Oxana [17]
3 years ago
11

LB. Johnson middle school

Mathematics
1 answer:
Wewaii [24]3 years ago
5 0

Answer:

What is this????

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Mr North spent $144,00 to build a fence around the perimeter of bis vegetable garden.He paid $6.oo per yard for fencing a, Draw
DedPeter [7]
The perimeter of his fence is 2400 the steps i took to find this was i divided 14400 by 6 an i got 2400.  I can only think of one possible reason I hope the helps, Sorry
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3 years ago
How do you do rotation of the origin ?
anzhelika [568]
The rotation rule would be (-y, x)

Write/mark all your coordinates down. Now plot all your prime points and draw a line connecting them.

Hope this helps!~
6 0
3 years ago
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drek231 [11]
Ok it will be c so yea
4 0
2 years ago
If the roots of a quadratic equation are 1±√5, then the product of the roots is
Drupady [299]
The plus-minus sign represents that there are two possible outcomes.

In this case, we have 1 \pm \sqrt{5}. When we branch out the possibilities we got 2 values: 1 + \sqrt{5} and 1 - \sqrt{5}

Those are the roots of this equation. When they ask their product, they want you to multiply both numbers.

When we multiply them: (1 + \sqrt{5}) \times( 1 - \sqrt{5})

When we FOIL the we get: 1 \times 1 - 1 \times \sqrt{5} + 1 \times \sqrt{5} - \sqrt{5} \times \sqrt{5}

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5 0
3 years ago
Read 2 more answers
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
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