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Andrei [34K]
3 years ago
5

What value of h makes the statement true? (17+h)+(–9) = 17+(5+(–9))

Mathematics
1 answer:
aalyn [17]3 years ago
5 0
I know this is late and you probably got the answer but just in case its C. 5. :P
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The endpoints of CD are C(1, -6)and D(7,5). Find the coordinate of the midpoint, M
yuradex [85]

Answer:

(4, - 0.5 )

Step-by-step explanation:

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

[ \frac{x_{1}+x_{2}  }{2} , \frac{y_{1}+y_{2}  }{2} ]

Here (x₁, y₁ ) = C(1, - 6) and (x₂, y₂ ) = D(7, 5) , then

midpoint = ( \frac{1+7}{2} , \frac{-6+5}{2} ) = \frac{8}{2} , \frac{-1}{2} ) = (4, - 0.5 )

6 0
3 years ago
Round-Trip Cost ($) Departure City Atlanta Salt Lake City Cincinnati 340.1 570.1 New York 321.6 354.6 Chicago 291.6 465.6 Denver
Lemur [1.5K]

Answer:

\bar x = 356.73 ---Atlanta

\bar x = 400.95 --- Salt Lake City

Step-by-step explanation:

Given

The round trip flight data of Atlanta and Salt Lake City

Required

The mean price of both

Mean is calculated as:

\bar x = \frac{\sum x}{n}

For Atlanta

\bar x = \frac{291.6 + 321.6 + 340.1 + 339.6 + 359.6 + 384.6 + 309.6 + 415.6 + 293.6 + 249.6 + 539.6 + 455.6 + 359.6 +333.9}{14}

\bar x = \frac{4994.2}{14}

\bar x = 356.73

For Salt Lake City

\bar x = \frac{570.1 + 354.6 + 465.6 + 219.6 + 311.6 + 297.6 + 471.6 + 618.4 + 513.6 + 523.2 + 381.6 + 159.6 + 267.6 + 458.6}{14}

\bar x = \frac{5613.3}{14}

\bar x = 400.95

5 0
3 years ago
My teacher never taught us this so idk how to do it
Verdich [7]

{x}^{3}  = 6\Leftrightarrow  \sqrt[3]{ {x}^{3} }  =  \sqrt[3]{6} \Leftrightarrow x =  \sqrt[3]{6}

6 0
3 years ago
What are the roots of the polynomial equation x3 - 10x=-3x2+ 24? Please help
snow_tiger [21]

Answer:

x = -2, 3, -4.

Step-by-step explanation:

x^3 - 10x = -3x2 + 24 = 0

x^3 + 3x^2 - 10x - 24 = 0

The last term is 24 and the coefficient of x^3 is 1 so +/- 1,  +-2, +/- 3 and +/- 4 could be among the roots ( by The Rational Root Theorem).

Let x = 1

f(x) = 1^3 +3(1)^2 - 10 - 24 = -30 so it's no t .

f(-1) = -1 + 3 + 10 - 24 = -12  so it's not -1.

Let  x = 2:

f(2) =  2^3 +3*4 - 20 - 24 = -32  so its not 2

f(-2) = -8 + 12 + 20 - 24 = 0  so x = -2 is a root

and therefore x+ 2 is a factor and we divide:

x + 2 ) x^3 + 3x^2 - 10x - 24   (  x^2 + x - 12   <------ The quotient

          x^3 + 2x^2

                      x^2 - 10x

                       x^2 + 2x

                              -12x - 24

                              -12x - 24

                                ..............

Now x^2 + x - 12 = (x - 3)(x + 4)

So (x + 2)(x - 3)(x + 4) = 0

This gives

x = -2, 3, -4.

7 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%5Clarge%5Cbegin%7Bbmatrix%7D%20%5Cbegin%7Barray%7D%20%7B%20l%20l%20%7D%20%7B%202%20%7D%20%
SVETLANKA909090 [29]

\huge \boxed{\mathbb{QUESTION} \downarrow}

\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}

In matrix multiplication, the number of columns in the 1st matrix is equal to the number of rows in the 2nd matrix.

\left(\begin{matrix}2&3\\5&4\end{matrix}\right)\left(\begin{matrix}2&0&3\\-1&1&5\end{matrix}\right)

Multiply each element of the 1st row of the 1st matrix by the corresponding element of the 1st column of the 2nd matrix. Then add these products to obtain the element in the 1st row, 1st column of the product matrix.

\left(\begin{matrix}2\times 2+3\left(-1\right)&&\\&&\end{matrix}\right)

The remaining elements of the product matrix are found in the same way.

\left(\begin{matrix}2\times 2+3\left(-1\right)&3&2\times 3+3\times 5\\5\times 2+4\left(-1\right)&4&5\times 3+4\times 5\end{matrix}\right)

Simplify each element by multiplying the individual terms.

\left(\begin{matrix}4-3&3&6+15\\10-4&4&15+20\end{matrix}\right)

Now, sum each element of the matrix.

\large\boxed{\boxed{\left(\begin{matrix}1&3&21\\6&4&35\end{matrix}\right) }}

7 0
2 years ago
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