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3 years ago

What is the degree A. 2 B. 3 C. 6 D. 8

Mathematics

1 answer:

sasho [114]3 years ago

4 0

6 hope i could help good luck

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QUESTION 5 What is the area of the trapezoid?

mr Goodwill [35]

Answer:

126 in^2

Step-by-step explanation:

The area of a trapezoid is found by

A = 1/2 (b1+b2) h

A = 1/2 ( 12+16)*9

A = 1/2 (28) *9

= 126

7 0

4 years ago

Read 2 more answers

Help with this one!!!<br><br><br> Show steps please:)

Ymorist [56]

<h3>Answer: 7.20</h3>

Work Shown:

18% = 18/100 = 0.18

18% of 40 = 0.18*40 = 7.20

4 0

3 years ago

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Express it in slope-intercept form.

motikmotik

Answer:

y = 1/4x - 4

Step-by-step explanation:

The slope is four points behind the origin, so we subtract -4 from the equation:

y = 1/4x - 4

7 0

3 years ago

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Leah invested $950 in an account paying an interest rate of 1.5% compounded continuously. Assuming no deposits or withdrawals ar

Contact [7]

$\bf ~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$950\\ r=rate\to 1.5\%\to \frac{1.5}{100}\dotfill &0.015\\ t=years\dotfill &6 \end{cases} \\\\\\ A=950e^{0.015\cdot 6}\implies A=950e^{0.09}\implies A\approx 1039.5\implies \stackrel{\textit{rounded up}}{A=1040}$

6 0

3 years ago

Solve the following matrix equations: (matrices)

Masja [62]

Step-by-step explanation:

$3X + \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} - 1 & 6 \\ 10 & 14 \end{pmatrix} \\ \\ 3X = \begin{pmatrix} - 1 & 6 \\ 10 & 14 \end{pmatrix} - \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} \\ \\ 3X = \begin{pmatrix} - 1 - 2 & 6 - 3 \\ 10 - 4 & 14 - 5 \end{pmatrix}\\ \\ 3X = \begin{pmatrix} - 3 & 3 \\ 6 & 9\end{pmatrix}\\ \\ X = \frac{1}{3} \begin{pmatrix} - 3 & 3 \\ 6 & 9\end{pmatrix}\\ \\ X = \begin{pmatrix} \frac{ - 3}{3} & \frac{3}{3} \\ \\ \frac{6}{3} & \frac{9}{3} \end{pmatrix}\\ \\ \huge \red{ X} = \purple{ \begin{pmatrix} - 1 &1 \\ 2 & 3 \end{pmatrix}}$

$3X + 2I_3=\begin{pmatrix} 5 & 0 & -3 \\6 & 5 & 0\\ 9 & 6 & 5\end{pmatrix} \\\\3X + 2\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\6 & 5 & 0\\ 9 & 6 & 5 \end{pmatrix} \\\\3X + \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} \\\\3X =\begin{pmatrix} 5 & 0 & -3 \\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \\\\3X =\begin{pmatrix} 5-2 & 0-0 & -3-0 \\ 6-0 & 5-2 & 0-0 \\ 9-0 & 6-0 & 5-2 \end{pmatrix} \\\\3X =\begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X =\frac{1}{3} \begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X =\begin{pmatrix} \frac{3}{3} & \frac{0}{3} & \frac{-3}{3} \\\\ \frac{6}{3} & \frac{3}{3} & \frac{0}{3} \\\\ \frac{9}{3} & \frac{6}{3} & \frac{3}{3} \end{pmatrix} \\\\\huge\purple {X} =\orange{\begin{pmatrix} 1 & 0 & - 1\\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}}\\$

8 0

3 years ago