<span>the product lies between 3 *3/4</span>
Answer:
h = -6
Step-by-step explanation:
Use the quadratic formula
ℎ
=
−
±
2
−
4
√
2
h=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}
h=2a−b±b2−4ac
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.
ℎ
2
−
3
6
=
0
h^{2}-36=0
h2−36=0
=
1
a={\color{#c92786}{1}}
a=1
=
0
b={\color{#e8710a}{0}}
b=0
=
−
3
6
c={\color{#129eaf}{-36}}
c=−36
ℎ
=
−
0
±
0
2
−
4
⋅
1
(
−
3
6
)
Answer:
Because if you were to switch 3-5 to 5-3 the answer would change. Plz mark me brainliest
Step-by-step explanation:
Answer: The required solution is

Step-by-step explanation: We are given to solve the following differential equation :

Let us consider that
be an auxiliary solution of equation (i).
Then, we have

Substituting these values in equation (i), we get
![m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmt%7D%2B10me%5E%7Bmt%7D%2B25e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E2%2B10y%2B25%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B10m%2B25%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B2%5Ctimes%20m%5Ctimes5%2B5%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B5%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-5%2C-5.)
So, the general solution of the given equation is

Differentiating with respect to t, we get

According to the given conditions, we have

and

Thus, the required solution is

The line DB is 8cm which is made up of DE and EB which are equal to each other, thus DE and DB are both 4 cm.
Secondly, since AC and DB are perpendicular, that means angle DEC and angle DEA are ninety degrees.
Thus using the Pythagorean theorem, we can find the length of AE and EC which we can add up to find the length of AC.

By adding AE and EC, we get that the length of AE and EC is 7.47 or as approximated to the nearest tenth 7.5.
Hope that helps!