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3 years ago

11

Let h be the actual height of a building. The height was measured as 47.7 m. Complete the following inequality. StartAbsoluteVal

ue h minus 47.7 EndAbsoluteValue less than or equals___

1 answer:

Orlov [11]3 years ago

7 0

Answer:

Step-by-step explanation:

|h-47.7|≤0

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-3 1/6 + 2 7/9 + 5 2/3 estimated

adelina 88 [10]

Answer:

5.28 if you round to the nearest hundredth. Drop a best answer and a like if it helped

3 0

3 years ago

Find the perimeter and area of the figure. Round to the nearest tenth if necessary. 33.7 mm; 108 mm2 33.7 mm; 54 mm2 30.7 mm; 50

Deffense [45]

Answer:

Perimeter: 33.7 mm

Area: 54 mm

Step-by-step explanation:

If you do 12 + 9.7 + 12 you will get the perimeter, and to get the area you would use the formula A = $\frac{1}{2}$ x base x height. Since the base is 12 and the height is 9, you would do 12 x 9 x 0.5 to get 54 mm for the area of the triangle.

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3 years ago

The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

Let a matrix A= $\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]$ and B= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

$\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0$

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

$\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]$

Apply $R_2\rightarrow R_2-R_1$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]$

Apply $R_3\rightarrow R_3+R_2$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]$

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

$\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Apply $R_1\rightarrow R_1-R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

4 0

3 years ago

5 out of 8 bags of m and m's had exactly 32 pieces in them. How many bags out of 40 should have exacty 32 pieces

Andre45 [30]

the answer i think is 64

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3 years ago

Arlo invested $4000 in an account that earns 5.5% interest, compounded annually. The formula for compound interest is A(t)=P(1+i

USPshnik [31]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$4000\\ r=rate\to 5.5\%\to \frac{5.5}{100}\dotfill &0.055\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases} \\\\\\ A=4000\left(1+\frac{0.055}{1}\right)^{1\cdot 4}\implies A=4000(1.055)^4\implies A\approx 4955.2986$

4 0

3 years ago

Read 2 more answers