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leonid [27]
3 years ago
7

Convert the function into intercept form. Show your work. y=-x^2+5x+36

Mathematics
1 answer:
UkoKoshka [18]3 years ago
7 0
Y = -x2 + 5x + 36 <span>→ y = -(x2 -5x -36)


</span><span>→ y =  -(x2 - 9x +4x - 36)


</span><span>→ y = -[x(x-9) + 4(x - 9)]


</span><span>→ y = -(x - 9)(x + 4)

Your answer would be </span>y=-(x-9)(x+4).
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Holden drove 110 miles in 2 3/4 hours. What was Holden's speed in miles per hour?
bazaltina [42]

Answer:

Holden's seed is 40 miles per hour

Step-by-step explanation:

The given parameters are;

The distance Holden drove = 110 miles

The  time duration in which Holden drove = 2 and 3/4 hours = 11/4 hours = 2.75 hours

Therefore, Holden's speed, s, can be found from the relationship;

Speed, s = Distance/time = 110 miles/(11/4 hours) = 40 miles per hour

Holden's seed = 40 miles per hour.

8 0
3 years ago
Write an equation for the
Thepotemich [5.8K]

Answer:

12+3x=27

Step-by-step explanation:

Follow the words and write it as you read it, if it's (times) you multiply and if it's (larger), it's addition.

5 0
3 years ago
Read 2 more answers
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zaharov [31]
It's hard to explain by words so sorry if you don't understand. For the graphing part, starting from (-3, 0), plot points in (1, 2), (6, 3), and connect them. The end of a line should be (10, 3.5). You will understand better if you use graphing calculator.

The answer of a second question is {x|x is greater or equal to -3.
7 0
3 years ago
Х<br> If<br> 5.3<br> =<br> 12, what is the value of x?<br> 8<br> 0 ООО<br> 18<br> o 24
gladu [14]

Answer:18

Step-by-step explanation:

8 0
3 years ago
) find a vector parallel to the line of intersection of the planes 5x − y − 6z = 0 and x + y + z = 1.
snow_tiger [21]
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.

Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>

We calculate the cross product as a determinant of (i,j,k) and the normal products

    i   j   k
   5 -1 -6
   1  1  1

=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>

Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0

Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
5 0
3 years ago
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