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3 years ago

5

A store has 300 televisions on order, and 30% are high definition. How many televisions on order are high definition? You may fi

nd the bar model useful in completing this problem.

1 answer:

scZoUnD [109]3 years ago

3 0

Answer:90

Step-by-step explanation:

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Five to to the 5th power over five to the eight

sveticcg [70]

$\frac{5^{5} }{5 ^{8} }$
Hope I helped, also, I used the math thingy, which I haven't used before, so I'm not sure quite how this turned out, well, I guess we will see in a minute! But if it does not work I can try it just typing regularly.

8 0

3 years ago

Hellllllppppppppppppppppppppppppppppp

larisa [96]

Answer:

c

Step-by-step explanation:

7 0

3 years ago

trasher [3.6K]

Tan60=x/12 12tan60=14.3 so area formula: 85.8

5 0

3 years ago

PICTURE^^ PLZ HELP !!!

densk [106]

QUESTION A

The solutions to the system of equations are the point of intersection of the quadratic graph and the straight line.

The two curves intersected at two points. Hence there are two real solutions.

The solutions are (-3,-1) and (-6,5).

QUESTION B.

The straight line is a tangent to the quadratic graph.

This means that the straight line intersected the quadratic graph at only one point.

The system has only one real solution.

The solution is (4,3).

QUESTION C

The quadratic graph and the straight line intersected at two distinct points.

The system has two real solutions.

The solutions are (0,3) and (6,-3).

QUESTION D

The two graphs are hanging apart.

There is no point of intersection.

The system has no real solutions.

QUESTION E

The straight line and the quadratic curve intersected at two distinct points.

The system has two real solutions.

The solutions are (0,0) and (1,1).

QUESTION F

The two curves intersected at only one point.

There is only one real solution to the system.

The solution is (-1,5).

3 0

3 years ago

The height, h, in meters above the ground, of a projectile at any time, t, in seconds, after the launch is defined by the functi

Ivanshal [37]

The equation $h(t) =-6t^2 + 15t + 2$ <span>sets the parabola. You can find the parabola vertex as:
</span>
$t_v= \frac{-b}{2a} =- \frac{15}{2\cdot (-6)} = \frac{15}{12}= \frac{5}{4}=1.25$

then

$h( \frac{5}{4} )=-6\cdot (\frac{5}{4} )^2+15\cdot \frac{5}{4} +2= \frac{91}{8} =11.375$

This means that the maximum height is 11.375 m and <span>a projectilereaches this height at 1.25 seconds.</span><span /><span>
</span><span>
</span><span>The projectile initially launched at time t=0, so the initial height is h(0)=2 m</span>

6 0

4 years ago