All categories

3 years ago

End SAYS 130 HELPPPP

Mathematics

2 answers:

Mashcka [7]3 years ago

6 0

B that is your correct answer

vivado [14]3 years ago

3 0

Answer:

it might be d

Step-by-step explanation

You might be interested in

Whats the volume of the prism please show work‼️

attashe74 [19]

Answer:

Step-by-step explanation:

Volume of a prism:

Area of cross section x height

4 0

2 years ago

Please help with this

slava [35]

Answer:

D) x= $\frac{-2+-2\sqrt{7} }{3}$

Step-by-step explanation:

The quadratic formula is : $\frac{-b+-\sqrt{b^2-4ac} }{2a}$

A in terms of this question=9

B in terms of the question is 12

C in terms of the question is -24.

This question is an example of a quadratic equation. To work this out you may first need a calculator. The first step is to substitute the values of a,b and c into the formula. So once substituted the formula of $\frac{-b+-\sqrt{b^2-4ac} }{2a}$ becomes $\frac{(-12)+-\sqrt{(12)^2-4*(9)*(-24)} }{2(9)}$ . Although when written in a calculator there will not be a plus and minus button and so you would have to do this separately.

However when substituting the values it would be best practice to put them in brackets.

1) Substitute the values into the equation for +.

$\frac{(-12)+\sqrt{(12)^2-4*(9)*(-24)} }{2(9)}=\frac{-2+2\sqrt{7} }{3} /1.097167541$

2) Substitute the values into the equation for -.

$\frac{(-12)-\sqrt{(12)^2-4*(9)*(-24)} }{2(9)}=\frac{-2+2\sqrt{7} }{3} /-2.430500874$

8 0

3 years ago

Rule add 15, subtract 10 first term4

mr Goodwill [35]

I think that would be 9 I am not sure.

8 0

2 years ago

Read 2 more answers

A proof should always begin with stating the given information.<br> True<br> False

const2013 [10]

<span>A proof should always begin with stating the given information.
True</span>

5 0

3 years ago

Read 2 more answers

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