<span>The answers to this problem are:<span>(<span>±5</span></span>√3/8,±5/8)<span>Here is the solution:
Step 1: <span><span><span>x2</span>+<span>y2</span>=<span>2516</span>[2]</span><span><span>x2</span>+<span>y2</span>=<span>2516</span>[2]</span></span>
Step 2: Substitute:<span>
</span><span><span>8<span><span>(<span>25/16</span>)^</span>2</span>=25(<span>x^2</span>−<span>y^2</span>)
</span><span>8<span><span>(<span>25/16</span>)^</span>2</span>=25(<span>x^2</span>−<span>y^2</span>)</span></span>
</span><span>x^2</span>−<span>y^2</span>=<span>25/32</span><span>.
Add [2] and [3]:<span>
</span><span>2<span>x^2</span>=<span>75/32
</span><span>x^2</span>=<span>75/74</span></span>
<span>x=±5</span></span>√3/8<span>
Substitute into [2]:<span>
</span><span><span>75/64</span>+<span>y^2</span>=<span>50/32
</span><span>y^2</span>=<span>25/64</span></span>
<span>y=±<span>5/8</span></span>
</span>
</span>
Answer:
Ball hits the ground after 4.5 sec
Step-by-step explanation:
Let a -1, so that the leading coefficient is positive
So our quadratic is
The key coefficients of two binomial variables can be 1 and 16, or 2 and 8, or 4 and 4, for the leading coefficient of 16.
Yet they can't actually be 4 and 4 because the linear (x) term coefficient has to be a multiple of 4, which it isn't and leading coefficients 1 and 16 on the binomial factors is not likely.
So, 2 and 8 taken as the leading coefficients of two binomial factors.
For constant 405, possible factorizations are 
Taking first factor, thus we find negative value for given time t. But second time equivalent to zero gives the value of 4.5 for t
Thus ball hits the ground after 4.5 sec
.
First one:
you can add -10m and -13m but you can't add -10m and 2m^4 becuase the powers aren't the same so
when adding the like terms
look at the:
powers, (x^3 adds with x^3)
placehloder letter (x adds with x and y adds with y and so on)
-10m+2m^4-13m-20m^4
powers: m^1 and M^4
placeholders: all m
add
-10m-13m+2m^4-20m^4
-23m-18m^4
second one:
when multiplying exponents, you add with like
so if you multipliy
x^2yz^3 times x^4y^2z^2 thne you would get x^6y^3z^5
when multiply with coeficients
2x^2yz^3 times 4x^4y^2z^2=8x^6y^3z^5
so using associative property a(bc)=(ab)c
2/3 times p^4 times y^3 times y^4 times s^5 times 6 times p^2 times s^3
group like terms
(2/3 times 6) times (p^4 times p^2) times (y^3 times y^4) times (s^5 times s^3)
(4) times (p^6) times (y^7) times (s^8)
4p^6y^7s^8
Answer:
If you are rounding to the nearest tens the smallest number is 45. If you are rounding to the nearest whole number, the smallest number is 49.5
Let us consider 48 as 100%
so,
48------------->100
x -------------->75
therefore,
100*x=48*75
x=(48*75)/100
x=36
