5.25% turns into 0.0525. <span>Then, you multiply 0.0525 and 1,000 together. This equals 52.50. </span><span>So, you would earn $52.50 at the end of the month in interest.</span>
The circle is divided in to 6 pieces.
one of the pieces is shaded.
The shaded portion/the number of pieces
1/6
\left[a _{3}\right] = \left[ \frac{ - b^{2}}{6}+\frac{\frac{ - b^{4}}{3}+\left( \frac{-1}{3}\,i \right) \,\sqrt{3}\,b^{4}}{2^{\frac{2}{3}}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{24}+\left( \frac{1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{\sqrt[3]{2}}\right][a3]=⎣⎢⎢⎢⎢⎡6−b2+2323√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))3−b4+(3−1i)√3b4+3√224−3√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))+(241i)√33√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))⎦⎥⎥⎥⎥⎤
Answer:
9/2
Step-by-step explanation:
this is a simple integral function
given limits of interval [a,b] of a continuous function f(t), you can find the area under the curve by using:
using the fundamental theorem of calculus that states the integral of f(x) in the interval [a,b] is = g(a)-g(b), where g(x) is the antiderivative of f(x)
our g(x) =
g(3)-g(0) = g(3) = 27/2 - 27/3 = 27/2-9 = 9/2
Answer:
2:3 is
✔ less than
5:6
10:12 is
✔ greater than
6:9
The ratios in Table 1 are
✔ less than
the ratios in Table 2
Step-by-step explanation: