All categories

2 years ago

A triangle has a side lengths of 34in., 28in., and 42in. is the triangle an acute, right, or obtuse?

Mathematics

2 answers:

Ghella [55]2 years ago

4 0

Answer:

Acute Triangle

An acute-angled triangle is a type of triangle in which all the three internal angles of the triangle are acute, that is, they measure less than 90°

Step-by-step explanation:

Hope this helps:)

jenyasd209 [6]2 years ago

3 0

It would be an acute

You might be interested in

The perimeter of the kite is 134 cm. Find the value of x.

frez [133]

Answer:

x = 10 cm

Step-by-step explanation:

(3x + 2) + (3x + 2) + (4x - 5) + (4x - 5) = 134 cm

combine like terms:

14x - 6 = 134

add 6 to each side of the equation:

14x = 140

divide both sides by 14:

x = 10

3 0

3 years ago

Write in point slope form an equation of the line that passes through the given point and has the given slope: (0,1); m= 2

sergey [27]

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

$\diamond\large\blue\textsf{\textbf{\underline{\underline{Question:-}}}}\diamond$

Write an equation for the line that passes through (0, 1) and has a slope of 2 (in point-slope form).

$\diamond\large\blue\textsf{\textbf{\underline{\underline{Answer and How to Solve:-}}}}\diamond$

<u>Point-slope form</u>:-

$\sf{y-y_1=m(x-x_1)}$

Substitute 1 for y₁, 2 for m, and 0 for x₁:-

$\sf{y-1=2(x-0)}$

So we conclude that Option B is correct.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

3 0

2 years ago

PLEASE ANSWER!!! A certain transformation moves a line segment as follows: Pt. A (2, 1) moves to A' (2, -1) and Pt. B (5, 3) to

erma4kov [3.2K]

Answer:

Step-by-step explanation:

eee

3 0

3 years ago

Read 2 more answers

The volume of a triangular prism with base dimensions of 4 cm height and 10 cm base and a prism height of 15 cm is??

jonny [76]

I think you can do this way:
Volume = B*h where B is the area of the base.
B= 4 * 10 = 40
V= 40 * 15
V= 600 cm^3

5 0

3 years ago

Read 2 more answers

Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.

vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

$cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}$

$\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}$

4 0

2 years ago