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4 years ago

Can I get some help with this

Mathematics

1 answer:

Arturiano [62]4 years ago

6 0

I think it would be 4.7A - C

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HELP PLEASE THANK YOU

pochemuha

Answer:

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8 0

3 years ago

PLEASE HELP ASAP ON MY MATH!!

Allisa [31]

Answer:

Area of shaded region=41 square feet

Area of non shaded region= 87 square feet

Step-by-step explanation:

As, Shaded area is made of different shapes including two rectangles and one triangle

So,

Area of shaded region,

$=(8\times2)+(10\times 1)+(\frac{1}{2}\times 3\times10)\\\\ =16+10+15=41 \quad \text{square feet}$

Area of non shaded region=total area of rectangular region-Shaded region

$=(16\times 8)-(41)\\=128-41\\=87 \quad \text{square feet}$

7 0

2 years ago

Read 2 more answers

DFG and JKL are complementary angles. M DFG = x+5 and m JKL=x-9. The measure of DFG is ___, and the measure of JKL is ___.

jeyben [28]

Answer:

Step-by-step explanation:

the sum of complementary angles = 90°, hence

x + 5 + x - 9 = 90

2x - 4 = 90 ( add 4 to both sides )

2x = 94 ( divide both sides by 2 )

x = 47

Thus ∠DFG = 47 + 5 = 52° and ∠JKL = 47 - 9 = 38°

6 0

3 years ago

Read 2 more answers

What is the equation of a line, in general form, that passes through point (1, -2) and has a slope of 1/3.

Studentka2010 [4]

Hello,

y+2=1/3*(x-1) ==>3y+6=x-1==>x-3y-7=0

Answer C

8 0

3 years ago

Read 2 more answers

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