The area of a square garden is 87 square meters. Estimate the side length of the garden 12 m 9 m 7 m 11 m
7) -(11)/(4)n+(1)/(4)n=-(11)/(2)
Choose all the true statements. A. sqrt (-2)=sqrt (2) B sqrt (-2)=sqrt (2i) C. sqrt (-8)=2isqrt (2) D. sqrt (-8)=-2sqrt (2) E. sqrt (-16)=4i 1F. sqrt (-16)=-4
2. angle JZQ=(9x+9)^circ and angle TLN=(8x+19)^circ
Solve the equation. vert x+5vert =3x+3 x=[?]
Simplify: (2-4i)(3-6i) A -18-24i B 6-24i C 6-12i (D) -18-12i
Which expression is equivalent to 6g+5-2g-4 Answer 4g+1 4g+9 8g+1 11g-6
Simplify the expression: 3(2x-1)-3x 6x-3 6x-1 2x+3 3x-3
lim _(xarrow 0)(7x^3-x^2)/(x)
(4)/(13)(x-3)-(15)/(13)=-5
Apply the distributive property to the expression. $100(0.07a+0.02b)=\square $
51. Find the volume of a pyramid with a square base, where the perimeter of the base is 15.8 ft and the height of the pyramid is 20.4 ft. Round your answer to the nearest tenth of a cubic foot.
4) $v^{1}=$ A) 0 B) 1 OC) v OD) $1/v$
What is the range of the function graphed above? $-\infty \lt y\leqslant 2$ $-\frac {3}{2}\leqslant y\lt \infty $ $-\infty \lt y\lt \infty $ $\{ 2\} $
2. You are inscribing a regular pentagon in a cIrcle. You need to know the angle measure between the vertices. You divide $360^{\circ }$ by what number? 3 s 6
Which equation represents a line that has a slope of $-\frac {1}{2}$ and passes through point $(4,-5)$ $y=-\frac {1}{2}x+\frac {3}{2}$ $y=-\frac {1}{2}x+\frac {13}{2}$ $y=-\frac {1}{2}x-7$ $y=-\frac {1}{2}x-3$
Smartphones: A poll agency reports that $38\% $ of teenagers aged $12-17$ own smartphones. A random sample of 120 teenagers is drawn. Round your answers to at least four decimal places as needed Part: 0/6 Part 1 of 6 (a) Find the mean $\mu _{\hat {p}}$ The mean $\mu _{\hat {p}}$ is 45.6000 Part: 1/6 Part 2 of 6 (b) Find the standard deviation $\sigma _{\hat {p}}$ The standard deviation $\sigma _{\hat {p}}$ is $\square $
1. If $x+3=7$ then x equals: A. 3 B. 4 C. 5 D. 10
Find the angle $\beta $ between $0^{\circ }$ and $180^{\circ }$ that satisfies the following equation. $21^{2}=20^{2}+29^{2}-(2)(20)(29)cos\beta $ $\beta \approx \square ^{\circ }$ (Round to one decimal place as needed.)
Solve the equation given by completing the square. $5x^{2}+20x-25=0$ [Hint: Divide by 5 first] $x=\square $