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2 years ago

So i need help with this pls i suck at algebra

Mathematics

1 answer:

Ronch [10]2 years ago

5 0

Answer:

The 5x^2 vs -5x^2 will reflect over "X" axis

the +1 vs -2 will shift the graph down three units

the first answer is the correct answer

Step-by-step explanation:

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Order of operations- usually, one egg is released each month during a female's reproductive years. If menstruation begins at 12

Art [367]

You have to subtract 12 from 50 to get 38 and multiply it by 12 (number of months) to get 456 eggs

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3 years ago

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Match each set of vertices with the type of quadrilateral they form.

Serjik [45]

Only correct the first one.

Please, see the graph in the attached file.

1) a rectangle with    A(3,3), B(3,6),
noncongruent ←→ C(7,6), D(7,3)
adjacent sides
(in green)

2) a parallelogram
with A(2,0), B(3,2)
nonperpendicular ←→ C(6,3), D(5,1)
adjacent sides
(in blue)

3) a square ←→ A(3,3), B(2,5),
(in red) C(4,6), D(5,4)

4) a rhombus with       A(2,-2), B(3,0),
nonperpendicular    ←→ C(4,-2), D(3,-4)
adjacent sides
(in black)

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3 years ago

Find the value of 4x if x/2=3

Doss [256]

Answer:

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

x/2 = 3

x = 3×2

x = 6

Therefore

4x = 4 × 6

4x = 24

4 0

3 years ago

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Which fraction is equivalent to -7/8 1. 8/-7 2. -7/-8 3. 8/7 4. 7/-8

IceJOKER [234]

Answer:

Your answer is answer no 4 = 7/-8

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3 years ago

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Let v1 =(-6,4) and v2=(-3,6) compute the following what i sthe angle between v1 and v2

Agata [3.3K]

$\bf ~~~~~~~~~~~~\textit{angle between two vectors }
\\\\
cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies 
\measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right)\\\\
-------------------------------$

$\bf \begin{cases}
v1=\ \textless \ -6,4\ \textgreater \ \\
v2=\ \textless \ -3,6\ \textgreater \ \\
------------\\
v1\cdot v2=(-6\cdot -3)+(4\cdot 6)\\
\qquad \qquad 42\\
||v1||=\sqrt{(-6)^2+4^2}\\
\qquad \sqrt{52}\\
||v2||=\sqrt{(-3)^2+6^2}\\
\qquad \sqrt{45}
\end{cases}\implies \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{52}\cdot \sqrt{45}} \right)
\\\\\\
\measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{2340}} \right)\implies \measuredangle \theta \approx 29.74488129694^o$

8 0

3 years ago