All categories

3 years ago

Classify the four angles of the quadrilateral

Mathematics

2 answers:

salantis [7]3 years ago

7 0

Answer:

<A = right angle

<B = acute angle

<C = obtuse angle

<D = acute angle

Step-by-step explanation:

Right angle = 90 degrees

Acute angle = less than 90 degrees

Obtuse angle = more than 90 degrees

sattari [20]3 years ago

4 0

Answer:

hisbojsnjoabnonsiondokihsdih ;)

You might be interested in

A ball is launched straight up in the air from a height of 6 feet. Its velocity (feet/second) t seconds after launch is given by

german

Step-by-step explanation:

A ball is launched straight up in the air from a height of 6 feet. The velocity as a function of time t is given by :

f(t) = -32 t+285

Height of the ball is :

$h(t)=\int\limits{f(t){\cdot} dt}\\\\h(t)=\int\limits{(-32t+285){\cdot} dt}\\\\h(t)=-16t^2+285t+C$

C is constant. Here the ball is launched from a height of 6 feet. So,

$h(t)=-16t^2+285t+6$

At t = 2 s, $h(t)=-16(2)^2+285(2)+6=512\ m$

At t = 9 s, $h(t)=-16(9)^2+285(9)+6=1275\ m$

Between 2 s and 9 s, the ball's height changed is : 1275 - 512 = 763 m.

6 0

3 years ago

Help please with this question I will mark branliest

faltersainse [42]

Answer:The pictures not that clear

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Evaluate the expression when a=3 and b=24 b-4a

Inessa05 [86]

Answer:

12.

Step-by-step explanation:

$b - 4a$

Substitute this equation with a=3 and b=24:

$24 - 4(3)$

$24 - 12$

$12$

5 0

3 years ago

Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0

eduard

If you're using the app, try seeing this answer through your browser: brainly.com/question/2787701

_______________

$\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}$

Square both sides:

$\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}$

$\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\
\mathsf{17\,sin^2\,\theta=1}\\\\
\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}$

Since $\mathsf{sin\,\theta}$ is positive, you can discard the negative sign. So,

$\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}$

Substitute this value back into $\mathsf{(i)}$ to find $\mathsf{cos\,\theta:}$

$\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

7 0

3 years ago

9.6 *10 ^-4 in standard notation

8_murik_8 [283]

9.6 x 10^-4 = 0.00096

5 0

3 years ago