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I am Lyosha [343]
2 years ago
10

HELP NEEDED ASAP! GIVING BRAINLIEST

Mathematics
2 answers:
Tatiana [17]2 years ago
8 0
You know don’t count me on this but i believe it’s angle 7, because Congruent angles are angles with exactly the same measure.
Olin [163]2 years ago
7 0

Answer:

C (u=5)

B ( angle 4)

Step-by-step explanation:

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Street H is perpendicular to the line 2x –3y = –4 and goes (0, –6). Write this street’s equation in standard form.
storchak [24]

Answer:

I think its 68 + 1 = 69

Step-by-step explanation:

6 0
2 years ago
At top speed, a boat can cover 1,533 kilometers in 3 hours what unit rate does the boat travel during this time frame?
svlad2 [7]
The unit rate that the boat can travel during this time frame is 511 kilometers in 1 hour.

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4 0
3 years ago
Please help i'm so confused
umka21 [38]

Answer:

if it's bot raining, then it's sunny

8 0
3 years ago
27? i cant figure out her mistake can anyone help ?
expeople1 [14]
The mess up occurs in one of the first things she says. She first claims that 7^2 +25^2 = 24^2
Wait a minute... slow your roll there.
25 is the hypotenuse, or the longest side of the triangle AND opposite the right angle (but we aren't supposed to know that yet).
The Pythagorean Theorem holds that a^2 + b^2 = c^2, where a & b are the legs of the triangle & c is the hypotenuse. So, she got the formatting correctly, but she switched 24 & 25.
Hope this helps!
4 0
3 years ago
Read 2 more answers
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

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

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
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