What is the area of the triangle (-5,-2) (3,-1)

Mathematics

1 answer:

max2010maxim [7]3 years ago

3 0

idk

Step-by-step explanation:

im to tired this sucks

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Which is the biggest number 2.07 2.24 2.4 2 2.04

Temka [501]

Among all these no., see the first no. that lies before or left side of decimal point, here... 2 is common so next check the no. After decimal... 2.0, 2.2, 2.4, 2 is considered as 2.0 and then we have 2.0... Let's not take 2.0s and let's take 2.2 and 2.4 which is greater than 2.0. Next see the second no. That comes after decimal We have 2.24 and 2.4 which can be taken as 2.40... The no.s after decimal, 24 is greater than 40 so 2.4 is greater...

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3 years ago

What is the area of an isosceles trapezoid, if its shorter base has length of 18 cm, the length of the altitude is 9 cm, and the

padilas [110]

The formula for the area of a triangle of base b and altitude h is A = (1/2)(b)(h).

Here, b = 18 cm and h = 9 cm. Thus, the desired area is:

(18 cm)(9 cm)

A = ----------------------- = 81 cm^2

This problem gives you more info than is needed to solve it. The area of the given triangle is 81 cm^2.

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3 years ago

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$