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nataly862011 [7]
3 years ago
9

BEEN ASKING FOR AN HOUR PLEASE! a rectangle's area is 336 mm Its height is 16 mm. What is its bases?​

Mathematics
1 answer:
Fantom [35]3 years ago
7 0

Answer:

Base is 21

Step-by-step explanation:

I'm gonna be honest, just divide 336 by 16.

However just in case you meant volume instead of area because you said (bases), they would be 5 and 4.2

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Help?¿<br> Anyone know the answer to this and can show work?
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3 years ago
Graph the linear equation. Find three points that solve the equation, then plot on the graph.
Illusion [34]

Answers:

Three points that solve the equation: (1, \frac{19}{4} ) , (2, 5), (3, \frac{21}{4} )

The graph is shown in the attached pictures.

NOTE: The first picture is the graph of the equation along with the plotted points, and the second one shows the work for those three points.

Step-by-step explanation:

1. To graph this equation, an easier way to do it would be to convert to slope-intercept form so we can graph knowing the y-intercept and the slope. Do this by isolating the y on the left side like so:

x-4y=-18\\-4y = -x -18\\y = \frac{-x}{-4} -\frac{18}{-4} \\y = \frac{1}{4} x + \frac{9}{2}

Remember that slope-intercept form is in y = mx + b format, and that m is the slope and b is the y-intercept. With this information, we know that (0, \frac{9}{2}) is the y-intercept and \frac{1}{4} is the slope of this equation. We can plot the point (0, \frac{9}{2}) on the graph, and then use the slope of \frac{1}{4} from there to graph other points and form a line. (When I graphed the line, I didn't include these "other points" so it wasn't confusing to locate which points were the three solutions listed.)

2. Points that solve an equation - or solutions - are also points that the line of the equation intersects. So, what we can do is form a table, plug in some x values into the equation, and solve for a y-value. The x and y values will form a point that is on the graph, thus they are solutions. (Please look at the second picture for work and clarification.) After identifying these points, just plot them on the graph and label them (as shown in the first picture).

8 0
3 years ago
If an arrow is shot upward on Mars with a speed of 56 m/s, its height in meters t seconds later is given by y = 56t − 1.86t2.
Zina [86]

Answer: (a) (i)50.42 m/s (ii) 51.35 m/s (iii) 52.094 m/s (iv) 52.2614 m/s (v) 52.27814 m/s

(b) 54.14 m/s

Step-by-step explanation:

The average speed over the given time interval [a,b]: = \frac{y(b)-y(a)}{b-a}

Given: If an arrow is shot upward on Mars with a speed of 56 m/s, its height in meters t seconds later is given by y = 56t − 1.86t^2.

(a)

(i) average speed = \frac{y(2)-y(1)}{2-1}

=\frac{56(2)-1.86(2)^2-(56(1)-1.86(1)^2)}{1}

=50.42 m/s

(ii) average speed = \frac{y(1.5)-y(1)}{1.5-1}

=\frac{56(1.5)-1.86(1.5)^2-(56(1)-1.86(1)^2)}{0.5}

=51.35m/s

(iii) average speed = \frac{y(1.1)-y(1)}{1.1-1}

=\frac{56(1.1)-1.86(1.1)^2-(56(1)-1.86(1)^2)}{0.1}

=52.094m/s

(iv) average speed = \frac{y(1.01)-y(1)}{1.01-01}

=\frac{56(1.01)-1.86(1.01)^2-(56(1)-1.86(1)^2)}{0.01}

=52.2614&#10;m/s

(v) average speed = \frac{y(1.001)-y(1)}{1.001-01}

=\frac{56(1.001)-1.86(1.001)^2-(56(1)-1.86(1)^2)}{0.001}

=52.27814&#10;m/s

(b) y(1)=56(1)-1.86(1)^2

=56-1.86

= 54.14 m/s

3 0
3 years ago
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