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igomit [66]
3 years ago
6

Find the volume of triangular prism. NO LINKS

Mathematics
1 answer:
Brums [2.3K]3 years ago
7 0

Answer:

504

Step-by-step explanation:

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There are 6 dogs and 12 cats in the backyard. Write the ratio of cats to total animals In 3 ways, in simplest form.
Luda [366]

Step-by-step explanation:

12:18

12 to 18

12/18 = 2/3

5 0
3 years ago
What is the equation of the line that passes through point (4, 12) and has a y-intercept of —2?
zmey [24]

Answer:

<h2>y = 3.5x - 2</h2>

Step-by-step explanation:

The slope-intercept form of an equation of a line:

<h3>y = mx + b</h3>

m - slope

b - y-intercept

We have y-intercept b = -2. Substitute:

<h3>y = mx - 2</h3>

We have the point (4, 12) → x = 4, y = 12. Put them in the equation:

12 = 4m - 2         <em>add 2 to both sides</em>

14 = 4m       <em>divide both sides by 4</em>

14/4 = m → m = 3.5

5 0
3 years ago
Read 2 more answers
Help please&lt;3
ale4655 [162]
Y=-3/2x+4

The slope is -3/2, since Kerry traveled three feet and only went down 2.

The y intercept is 4, since she started four feet above the origin
3 0
3 years ago
What is the equation of this line in slope-intercept form (-1,5) and (1,-3)?
Harrizon [31]
Slope intercept means
y=mx+c
the slope of the line is m=(y2-y1)/(x2-x1)=(-3-5)/(1+1)= -4
the equation of the line is
y-5 = -4(x+1)
4x+y-1=0
changing the equation into slope intercept form
y= -4x+1
4 0
3 years ago
Is sin theta=5/6, what are the values of cos theta and tan theta?
mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill

\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}

5 0
3 years ago
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