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Leokris [45]
3 years ago
15

Help (look at picture)

Mathematics
1 answer:
mote1985 [20]3 years ago
8 0

Answer:

x is greater than or equal to -5 (the second graph)

Step-by-step explanation:

-5c+2<=27

-5x<=25

x>=-5

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On Sunday Lindsay rode her bikeb3 times as far as she did on Saturday. On Saturday she rode 3 miles. How far did she ride on Sun
Naddik [55]

Answer:

9 miles

Step-by-step explanation:

3 times 3 is 9

7 0
3 years ago
( m + 4 ) ( m + 1 )
love history [14]
(m+4)(m+1)

=m^2+5m+4........
3 0
3 years ago
Doug regularly mows his neighbor's lawn. Last month, Doug's neighbor paid him $55 for 6 hours of mowing. Is this situation most
nasty-shy [4]

Solution:

we are given that

Last month, Doug's neighbor paid him $55 for 6 hours of mowing.

we have been asked that

Is this situation most likely a proportional relationship?

yes..its as proportional relationship.

Because the amount that Doug get is $55 for 6 hours

It mean for each hour Doug get  is \frac{55}{6} $

It mean if you increase the number of hours then the number of dollars get paid also increases. Hence the given  situation is most likely a proportional relationship.



8 0
3 years ago
Solve −x+5≤4 or 2x+3≥−1 and write the solution in interval notation.
serg [7]

Answer:

Step-by-step explanation:

\left \{ {{-x+5\leq 4} \atop {2x+3\geq -1}} \right.   ⇒ x\geq 1\\2x\geq 4\\x\geq -2

or

x \geq -2\\(-2,+ { \infty} ) (Answer)

6 0
2 years ago
An object is propelled upward from the top of a 300 foot building. The path that the object takes as it falls to the ground can
serg [7]

Answer:

As per the statement:

The path that the object takes as it falls to the ground can be modeled by:

h =-16t^2 + 80t + 300

where

h is the height of the objects and

t is the time (in seconds)

At t = 0 , h = 300 ft

When the objects hit the ground, h = 0

then;

-16t^2+80t+300=0

For a quadratic equation: ax^2+bx+c=0         ......[1]

the solution for the equation is given by:

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

On comparing the given equation with [1] we have;  

a = -16 ,b = 80 and c = 300

then;

t= \frac{-80\pm \sqrt{(80)^2-4(-16)(300)}}{2(-16)}

t= \frac{-80\pm \sqrt{6400+19200}}{-32}

t= \frac{-80\pm \sqrt{25600}}{-32}  

Simplify:

t = -\frac{5}{2} = -2.5 sec and t = \frac{15}{2} = 7.5 sec

Time can't be in negative;

therefore, the time it took the object to hit the ground is 7.5 sec

8 0
3 years ago
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