Consider the following polynomial inequality: (1)/(2x+5)gt 1 Step 1 of 2:Write the polynomial inequality in the form p(x)lt 0,p(x)leqslant 0,p(x)gt 0,orp(x)geqslant 0 then find the real zeros of p(x)

Write the equation for the plane. The plane through the points $P(5,4,-36),Q(-3,6,-16)$ and $R(-1,-5,42)$ A. $4x+6y+z=8$ B. $4x+6y+z=-8$ C. $6x+y+4z=-8$ D. $6x+y+4z=8$

Evaluate the expression. $\frac {C(8,6)\cdot C(9,7)}{C(13,9)}$ $\square $

Simplify the expression: $-5(-6+2x)=$ $\square $

Find the area of the region that lies inside both curves. $r=3sin(\Theta ),\quad r=3cos(\Theta )$

Note: You may need to assume the fact that $\lim _{M\rightarrow \infty }M^{n}e^{-M}=0$ for all n. Decide whether or not the given integral converges. $\int _{0}^{\infty }e^{-5x}dx$ The integral converges The integral diverges. If the integral converges compute its value. (If the integral diverges enter DNE.) $\square $

What is the slope of the line that passes through the points $(-5,-9)$ and $(-5,-13)$ ? Write your answer in simplest form. Answer $\square $

Evaluar la expresión para $b=1.6$ $2\cdot b+3\cdot b$

Solve the equation, and check the solution. $\frac {1}{4}(3x+5)-\frac {1}{5}(x+7)=7$

Find an equivalent expression for $3csc(x-\frac {\pi }{2})$ using the cofunction identities. $3csc(x-\frac {\pi }{2})=\square $ (Simplify your answer.)

Use the trigonometric function values of quadrantal angles to evaluate the expression below $(cos180^{\circ })^{2}-(sin0^{\circ })^{2}$