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3 years ago

Some one please answer quick

Mathematics

1 answer:

azamat3 years ago

3 0

Answer:

Step-by-step explanation:

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Find the area of a square with a diagonal of 8cm.

klio [65]

You have to use the Pythagorean theorem but there is a little twist here since a square has has equal sides. So:
x^2+x^2=8
2x^2=8
2/2x^2=8/2
x^2=4
√x^2=√4
x=2
The side of the square is 2.
So now the area of a square is:
Side×Side
2×2=4
The area of the square is 4 centimetres

7 0

3 years ago

When 14 different second-year medical students measued teh systolic blood pressure of the same person, they obtained the results

malfutka [58]

Mean = 133.9
then calculate the z value
with z value you will find out the p value on a table.
if the p value is below .05 reject the null hypotheses

7 0

4 years ago

A planr leaves and flies due north at 280 miles per hour. AT the same time another plane takes off from the same airport and fli

kompoz [17]

Answer:

Distance between the two planes after 1 hour = 410 miles

Step-by-step explanation:

Given - A plane leaves and flies due north at 280 miles per hour. At

the same time another plane takes off from the same airport

and flies due east at a speed of 300 miles per hour.

To find - What is the distance between the two planes after 1 hour ?

Proof -

Given that , Plane 1 flies due north at a speed of 280 miles per hour.

So,

Total distance traveled in 1 hour = 280 miles

Also,

Plane 2 flies due east at a speed of 300 miles per hour.

So,

Total distance traveled in 1 hour = 300 miles.

The diagram is as follows :

Now,

We have to find the distance after 1 hour i.e. AC

In Triangle ABC ,

By using Pythagoras theorem, we get

AC² = BC² + AB²

⇒AC² = (300)² + (280)²

⇒AC² = 90,000 + 78,400

⇒AC² = 168,400

⇒AC = √168,400

⇒AC = 410.366 ≈ 410

∴ we get

Distance between the two planes after 1 hour = 410 miles

7 0

3 years ago

If two numbers on a fraction are negative, does that make the fraction positive

Sedbober [7]

I think the answer is yes

7 0

3 years ago

Suppose you have a water-balloon launcher. The balloon is 3 ft high when it leaves the launcher. Use the equation 0 = ?16t2 + 38

mario62 [17]

Answer:

t = 2.5 s

Step-by-step explanation:

Suppose you have a water-balloon launcher. The balloon is 3 ft high when it leaves the launcher. Its equation is :

$16t^2 + 38.8t + 3 =0$

The above equation is a quadratic equation. The general equation is :

$ax^2+bx+c=0$

Here, a = 16, b = 38.8 and c = 3

The solution of quadratic equation is given by :

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}\\\\t=\dfrac{-38.8\pm \sqrt{(38.8)^2-4\times (-16)\times 3} }{2\times (-16)}\\\\t=\dfrac{-38.8+\sqrt{(38.8)^{2}-4\times-16\times3}}{-2\times16}, \dfrac{-38.8-\sqrt{(38.8)^{2}-4\times-16\times3}}{-2\times16}\\\\t=-0.075\ s,2.5\ s$

So, at t = 2.5 s the balloon is in air.

8 0

3 years ago