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4 years ago

Find the point of intersection of the two lines y=3x+2 and y=2x-1

2 answers:

5 0

Basically the point of intersection is where the values of x and y are equal in both equations. In the first equation y is 3x+2. Substitute y in th second equation with 3x+2. The value of x is - 3. Do th same to get the value of y, which is - 7

inessss [21]4 years ago

4 0

(-3,-7) because the two lines have different slopes an the point that they intersect is the point i listed above

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Sin^2x/cos^2 + sinxcscx = sec^2x Identify it, please explain so I know how you did it, thanks

max2010maxim [7]

$\frac{sin^2x}{cos^2x} + sin x *cscx\\ tan^2x + \frac{sin x}{sin x} \\ tan^2x + 1\\ sec^2x$

$\frac{sin^2x}{cos^2x} = tan^2x$

$csc x = \frac{1}{sinx}$

$tan^2x + 1 = sec^2x$

6 0

3 years ago

Can someone give step by step on what to do?

ohaa [14]

Answer:

$SA=14.28^{2}$

Step-by-step explanation:

In the Equation it asks for 3 things; pi (3.14), r (The radius), and h (the height).

In the question, it says that the radius is 1 foot and that the height is 4 feet. From there all you need to do is insert those numbers where it's shown in the equation. This makes the equation go from this:

$SA=2\pi r^{2} + 2\pi r h$

To this: $SA=2(3.14)(1^{2}) + 2(1)(4)$

From there you follow PEMDAS/GEMDAS. And should have your answer.

7 0

3 years ago

The radius of a circle is 2.94cm. Which is closest to the area of the circle?

Marrrta [24]

Well you need to square 2.94 by 2 and then multiply by 3.14 to get your answer

5 0

3 years ago

Read 2 more answers

Please help me out with this

Naddik [55]

Answer:

(x - 5)² + (y + 3)² = 16

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

here (h, k) = (5, - 3) and r = 4, so

(x - 5)² + (y - (- 3))² = 4², that is

(x - 5)² + (y + 3)² = 16

3 0

3 years ago

Will give brainliest!

Gala2k [10]

Check the picture below, it hits the ground when y = 0.

$\bf ~~~~~~\textit{initial velocity}\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o \\\\
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{0}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{162}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+0t+162\implies 0=-16t^2+162\implies 16t^2=162
\\\\\\
t=\cfrac{162}{16}\implies t=\cfrac{81}{8}\implies t=10\frac{1}{8}$

6 0

3 years ago