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mamaluj [8]
3 years ago
15

If line y bisects AC, AB = 4-5x, and BC =2x+25, AC

Mathematics
1 answer:
jolli1 [7]3 years ago
7 0
4 - 5x = 2x + 25
4 = 7x + 25
-21 = 7x
-3 = x

4 - 5x + 2x + 25 = ac
4 - 5(-3) + 2(-3) + 25 = ac
4 - (-15) + (-6) + 25 = ac
19 + (-6) + 25 = ac
38 = ac
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HElp fasttttttttttttttttttttttttt
Eduardwww [97]

9514 1404 393

Answer:

  -0.16

Step-by-step explanation:

The 'a' value can be found by looking at the difference between the y-value of a point 1 unit from the vertex, and the y-value of the vertex.

Here, that is a negative fraction of a unit. If we assume the value is a rational number that can be accurately determined from this graph, then we can find it by looking for a point where the graph crosses a grid intersection. It looks like such grid points are (-7, 0) and (3, 0). The vertex is apparently (-2, 4), so the vertex form of the equation is ...

  y = a(x +2)^2 +4

Using the point (3, 0), we have ...

  0 = a(3 +2)^2 +4 . . . . . fill in the values of x and y

  -4 = 25a . . . . . . . . . . subtract 4; next, divide by 25

  a = -4/25 = -0.16

7 0
3 years ago
Which of the following is(are) the solution(s) to 16x-51- 7?
nexus9112 [7]
X= -2 because I love school soooo much
5 0
3 years ago
Read 2 more answers
A clothing company sells slacks that are either blue or gray and pleated or non-pleated. Last month, the company sold 5 times as
abruzzese [7]

Answer:

88

Step-by-step explanation:

5GP= BP

2BNP= GNP

BP+BNP=333

GP+GNP=225

BNP= 333-BP

BNP= 333-5GP

GP= 225-GNP

GP= 225-2BNP

BNP=333-5(225-2BNP)

BNP=333-1125+10BNP

-9BNP=-792

BNP=88

Hope it helps :)

Mark Brainliest if possible.

7 0
3 years ago
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Thompson High School had a population of 1800 students in 2010. Every year since then, the population has grown by 50 students a
Scorpion4ik [409]

I believe it would be B but I'm not 100% sure. Like i warn everyone, TRUST SOMEONE ELSE BEFORE TRUSTING ME. I would hate to see you get the wrong answer cause of me...

Best Hopes,

Cupkake~

3 0
3 years ago
Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
3 years ago
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