The height of a parallelogram is 5 feet The base is 7 feet the length is 9 1/2 feet what is the area of the parallelogram

Mathematics

1 answer:

Pie2 years ago

8 0

Answer:

[10 ft.]

Step-by-step explanation:

It's being given that there is a paralleolgram.

Let the length of its base is x ft.

Therefore, length of height = (x+6) ft.

Also, area of paralleolgram = 160 sq.

ft.

Now, we know that,

Area of paralleolgram = base x height

Therefore, we will get,

=> x(x+6) = 160

=> x² + 6x - 160 = 0

=> x² + 16x - 10x - 160 = 0

=> x(x+16) - 10(x+16)=0

=> (x+16) (x - 10) = 0

=> x +16=0 or X- - 10-0

=> x=-16

X = 10

But, we know that, length can't ever be negative.

Therefore, x = 10

Hence, length of base = 10 ft.

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An explosion causes debris to rise vertically with an initial speed of 120 feet per second. The formula h equals negative 16 t s

Novay_Z [31]

Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, $s=120\ ft/s$

The height 'h' of the debris above the ground is given as:

$h(t)=-16t^2+120t$

As per question, $h(t)=56\ ft$ . Therefore,

$56=-16t^2+120t$

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

$-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0$

Using quadratic formula to solve for 't', we get:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s$