Please help on this one please

Mathematics

1 answer:

borishaifa [10]3 years ago

5 0

The answer is a rhombus. A rhombus has for sides of equal shape, and the opposite sides a parrallel.

You might be interested in

Please answer this to the best of your ability

Lilit [14]

Answer:

For this exercise we need to solve the next equation:

3 +6x = 2x + 27

We can start substracting 3 at both sides:

3 + 6x - 2 = 2x + 27 - 3

6x = 2x + 25

And then, we can substract 2x::

6x - 2x = 2x + 24 - 2x

4x = 25

Then, we divide by 4:

4x/4 = 24/4

x = 24/4 = 6

Now, we can see if we have done it correctly:

3 + 6*6 = 39

2 * 6 + 27 = 39

And we have seen our result is correct

7 0

2 years ago

PLEASE HELP!!!!!!!!!!!!! DUE SOON

Sergio039 [100]

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t$

now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{65}{2(-16)}~~,~~0-\cfrac{65^2}{4(-16)} \right) \implies \left( \cfrac{65}{32}~,~0- \cfrac{4225}{-64}\right)$

$\bf \left( \cfrac{65}{32}~,~0+ \cfrac{4225}{64}\right)\implies \left( \stackrel{seconds}{2\frac{1}{32}}~~,~~ \stackrel{feet~hight}{66\frac{1}{64}}\right)$