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max2010maxim [7]
3 years ago
5

Which graph is generated by this table of values? x –4 0 3 y 1 2 3

Mathematics
1 answer:
Nookie1986 [14]3 years ago
7 0

Answer: See the graph attached.

Step-by-step explanation:

1. To solve this exercise you must plot the points given in the table of the problem, which are shown below:

(-4,1), (0,2), (3,3)

2. You must plot the first value of each ordered pair on the x-axis.

3. You must plot the second value of each ordered pair on the y-axis.

4. Therefore, when you plot them, you obtain the graph shown attached.

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An elephant went 119 km moving 42 km/h how many hours did the journey take?
Luden [163]

Answer:

1 HR ---> 42 km

? .----> 119 km

119 * 1/42

2.833

8 0
2 years ago
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Find the areas of the trapezoids.
IceJOKER [234]

Answer:

30 units^2

Step-by-step explanation:

The area of a trapezoid is

A = 1/2 (b1+b2)h

The scale is 2 units per line

b1 = 2 units

b2 = 8 units

h = 6 units

A = 1/2 ( 2+8) * 6

   = 1/2 (10)6 = 30 units^2

5 0
3 years ago
Question 8
sasho [114]

we are given

system of equations

First equation is

3x-2y=31

Second equation is

3x+2y=-1

now, we can find augmented matrix

A=\begin{pmatrix}3&-2&31\\ 3&2&-1\end{pmatrix}

now, we can change it into reduced row echelon form

step-1: multiply the 1st row by 1/3

A=\begin{pmatrix}1&-2/3&31/3\\ 3&2&-1\end{pmatrix}

step-2:add -3 times the 1st row to the 2nd row

A=\begin{pmatrix}1&-2/3&31/3\\ 0&4&-32\end{pmatrix}

step-3:multiply the 2nd row by 1/4

A=\begin{pmatrix}1&-2/3&31/3\\ 0&1&-8\end{pmatrix}

step-4: add 2/3 times the 2nd row to the 1st row

A=\begin{pmatrix}1&0&5\\ 0&1&-8\end{pmatrix}

so, we get

x=5,y=-8

Answer is (5,-8)

6 0
3 years ago
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PLEASE HELP ME WITH THIS I NEED IT
sukhopar [10]

Answer:

1400π m³

Step-by-step explanation:

V = BH

V = 10² * π * 14 = 1400π m³

5 0
3 years ago
Which of the following is the quotient of b and a?
vlabodo [156]
\bf 19^{\frac{7}{4}}\cdot \sqrt[a]{19^b}=19^{\frac{5}{2}}\sqrt{19}\\\\
-----------------------------\\\\
a^{\frac{{ n}}{{ m}}} \implies  \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-----------------------------\\\\
thus\qquad 19^{\frac{7}{4}}\cdot 19^{\frac{b}{a}}=19^{\frac{5}{2}}\cdot 19^{\frac{1}{2}}\implies 19^{\frac{7}{4}+\frac{b}{a}}=19^{\frac{5}{2}+\frac{1}{2}}
\\\\\\


\bf 19^{\frac{7}{4}+\frac{b}{a}}=19^{\frac{6}{2}}\implies 19^{\frac{7}{4}+\frac{b}{a}}=19^3\impliedby 
\begin{array}{llll}
\textit{same base, thus}\\
\textit{exponents must be the same}
\end{array}
\\\\\\
\cfrac{7}{4}+\cfrac{b}{a}=3\implies \cfrac{b}{a}=3-\cfrac{7}{4}
5 0
3 years ago
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