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Pepsi [2]
3 years ago
11

A car travels at a constant speed of 70 miles per hour. The distance the car travels in

Mathematics
2 answers:
HACTEHA [7]3 years ago
5 0

Answer:

huvfgyyrvfh

Step-by-step explanation:

Luba_88 [7]3 years ago
5 0
<h2><em>Answer: </em></h2><h2><em> </em></h2><h2><em>The distance function is  d(t)=70.t</em></h2><h2><em> </em></h2><h2><em>Step-by-step explanation: </em></h2><h2><em> </em></h2><h2><em>If an object moves at a constant rate of speed, we can determine how far the object travels by multiplying its rate by the time it has been traveling: </em></h2><h2><em> </em></h2><h2><em> distance=rate X time</em></h2><h2><em> </em></h2><h2><em>Rate is distance (given in units such as miles, feet, kilometers, meters, etc.) divided by time (hours, minutes, seconds, etc.). </em></h2><h2><em> </em></h2><h2><em>We know that a car travels at a constant speed of 70 miles per hour. </em></h2><h2><em> </em></h2><h2><em>So, the distance function is ,d(t)=70.t  where t is hours and the distance is in miles</em>.</h2>
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Sergio039 [100]

For a polynomial of the form ax^2+bx+c rewrite the middle term as a sum of two terms whose product is a⋅c=5⋅4=20 and whose sum is b=12.

<u>Factor 12 out of 12x.</u>

5x^2+12(x)+4

<u>Rewrite 12 as 2 plus 10</u>

5x^2+(2+10)x+4

Apply the distributive property.

5x^2+2x+10x+4

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

(5x^2+2x)+10x+4

Factor out the greatest common factor (GCF) from each group.

x(5x+2)+2(5x+2)

Factor the polynomial by factoring out the greatest common factor, 5x+25x+2.

(5x+2)(x+2)

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2 years ago
A square can be. Rhombus, and figure X is a rhombus, so figure X must be a square. True or false
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Read 2 more answers
Help please!!!!!!!!!
In-s [12.5K]

ANSWER

24


EXPLANATION

For a matrix A of order n×n, the cofactor C_{ij} of element a_{ij} is defined to be


   C_{ij} = (-1)^{i+j} M_{ij}


M_{ij} is the minor of element a_{ij} equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.


Here, we have


   C_{11} = (-1)^{1+1} M_{11} = M_{11}


M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.


   \begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right)  \end{aligned}


Since the determinant of a 2×2 matrix is


   \det\left(  \begin{bmatrix} a & b \\ c& d  \end{bmatrix} \right) = ad-bc


it follows that


   \begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \\ &= (0)(-8) - (-3)(8) \\ &= -(-24) \\ &= 24 \end{aligned}


so C_{11} = M_{11} = 24

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