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Montano1993 [528]
3 years ago
5

Describe and correct the error in the solution. (Be specific and explain as

Mathematics
1 answer:
sweet [91]3 years ago
6 0
The error occurred in step 2. You have to use the distributive property for the parenthesis. It is supposed to be 5p+15.

The correct solution is p>-13.
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I need an answer please! The question says Find the value of x that would make the following expression equivalent to sixteen ti
aev [14]
Square root of 6 times 16 will give us 39.19

The square root of 48 times 32 [which gives us 1536] is also equal to 39.19

The square root of 1536 is equal to 39.19

Hence, x is equal to 32

Hope this helped! :D
8 0
3 years ago
The formula equals 16t squared is used to approximate the distance​ s, in​ feet, that an object falls freely from rest in t seco
Brut [27]
The equation is actually h(t)=-16t^2+1437.  Free fall is always -16t^2 as the position function.  We are looking for how long it takes the object to hit the ground.  In other words, the height of an object is 0 when it is laying on the ground, so how long (t) did it take to get there? We will then set that position equal to 0 and solve for t.  0=-16t^2+1437.  If we subtract 1437 from both sides and divide by -16, we have t^2=89.8125.  Taking the square root of both sides gives us, rounded to the nearest tenth, t = 9.5 or t=-9.5.  The 2 things in math that will never EVER be negative are time and distance/length, so -9.5 is out.  That means that it took just about 9.5 seconds for the object to fall to the ground from a height of 1437 feet when pulled on by the force of gravity.
6 0
3 years ago
Simplify: (3x2y) · (5x3y2)
TiliK225 [7]

Answer:

15x^5*y^3

Step-by-step explanation:

6 0
3 years ago
The two triangles below are similar.
umka2103 [35]
I have a picture of the work for it but I don’t know how to upload it I can put a file maybe!
3 0
2 years ago
Sec²θ=4+2tan²θ<br>Find θ​
Marina86 [1]

Recall the Pythagorean identity,

\sec^2\theta=1+\tan^2\theta

So we have

\sec^2\theta=4+2\tan^2\theta

1+\tan^2\theta=4+2\tan^2\theta

-3=\tan^2\theta

But squaring any real number always gives a non-negative number, so there are no real solutions to this equation.

5 0
3 years ago
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