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Anna71 [15]
3 years ago
15

3. Molly bought four rose bushes for $16.40 each. Use

Mathematics
1 answer:
N76 [4]3 years ago
4 0

Answer:

$65.60

Step-by-step explanation:

4×16.40 = 4×(16+0.40) = (4×16) + (4×0.40) = 64 + 1.60 = 65.60

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What is the most accurate representation of the area of a plot measuring 11.7cm by 15.4cm ?
Aleks04 [339]
Area=width*height
area=(11.7)*(15.4)
area=180.18cm^2
7 0
3 years ago
Which inequality has a dashed boundary line when graphed? A y>=3/5x+1 B y>= -1/3x+1 C y>3x+1
Ket [755]

Answer: C y>3x+1

Step-by-step explanation:

  • When we graph an  inequality with strictly greater of less than sign ('<' or '>'), then the graph has a dashed boundary line .
  • Further it indicates that it does not include the points on the line.

From all the given options , only C contains inequality with '>' sign .

Hence, y>3x+1 is the inequality has a dashed boundary line when graphed.

hence, the correct option is C.

4 0
3 years ago
If the two terms of a gemotric sequence are a1=216, and a2=72, which is the third term? a3?
Ierofanga [76]
GEOMETRIC \: \: PROGRESSIONS \\ \\ \\\\Let \: the \: G.P. \: be \: \: A \: , \: Ar \: , \: A {r}^{2} \: ... \\ \\ Where \:first \: term \: is \: \: A \: \: \\ and \: common \: ratio \: is \: \: R \\ \\ Let \: An \: denotes \: the \: \: nth \: term \: of \: \\ the \: given \: Geometric \: Progression \: \\ \\ It \: is \: given \: - \\ \\ A1 \: = \: 72 \\ \\ A2 \: = \: 216 \\ \\ Common \: ratio \: = \: \frac{A2}{A1} = \frac{216}{72} \\ \\ R = 3 \\ \\ A \: = \: 72 \\ \\ A3 \: = \: A {r}^{2} = 72 \times 3 \times 3 \\ \\ \\ Hence \: , \: A3 \: = \: 648 \: \: \: \: \: \: \: Ans.
4 0
4 years ago
Name the method that is used to measure irregular shaped objects.
lys-0071 [83]

Answer:

Displacement

Step-by-step explanation:

hope this helps !

5 0
3 years ago
Solve the system of equations by transforming a matrix representing the system of equation into reduced row echelon form.
Gekata [30.6K]

Take the augmented matrix,

\left[\begin{array}{ccc|c}2&1&-3&-20\\1&2&1&-3\\1&-1&5&19\end{array}\right]

Swap the row 1 and row 2:

\left[\begin{array}{ccc|c}1&2&1&-3\\2&1&-3&-20\\1&-1&5&19\end{array}\right]

Add -2(row 1) to row 2, and -1(row 1) to row 3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&-3&4&22\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&9&36\end{array}\right]

Multiply through row 3 by 1/9:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&1&4\end{array}\right]

Add 5(row 3) to row 2:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&0&6\\0&0&1&4\end{array}\right]

Multiply through row 2 by -1/3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]

Add -2(row 2) and -1(row 3) to row 1:

\left[\begin{array}{ccc|c}1&0&0&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]

So we have \boxed{x=-3,y=-2,z=4}.

3 0
3 years ago
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