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Advocard [28]
3 years ago
15

9. What is the measure of BAC?

Mathematics
1 answer:
ddd [48]3 years ago
3 0

Answer:

20

Step-by-step explanation:

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Find the measure of each numbered angle.
Mamont248 [21]

Step-by-step explanation:

18z+10z+40z+5+1=360

68z+6=360

68z=360-6

68z=364

z≈5.353

7 0
2 years ago
Solve the system by substitution 3x-y=10 2x=y
Zanzabum

Answer

X=10 and y=20

Step-by-step explanation:

since y=2x, you need to substitute that into the equation to get:

3x-(2x)=10

And evaluate:

x=10

Then substitute 10 into the equation: 2x=y:

2(10)=y

evaluate:

20=y

Now go back to the original equation to check your answer and substitute the values in:

3(10)-(20)=10

30-20=10

10=10

6 0
3 years ago
Read 2 more answers
2) Choose the letter of the equation that best represents the problem. *
klemol [59]

Answer: x + 3x = 11

Step-by-step explanation:

Two cities A and B

Both cities receive 11 inches of rainfall in total

Amount of Rainfall in inches received by city A = x

Amount of Rainfall in inches received by city B = 3x

Therefore given the details above, total rainfall is best expressed using the equation ;

Rainfall in city A + Rainfall in city B = 11 inches

x + 3x = 11

3 0
3 years ago
Complete the following table by performing the operations to write equivalent systems.
Anvisha [2.4K]

Answer:

See below:

Step-by-step explanation:

Problem 1:

Multiply Equation 1 by 4, keep Equation 2 the same.

x+y=8, multiply each term by 4:

4*x=4x, 4*y=4y, 8*4=32

so, the equivalent system is: 4x+4y=32 and x-y=2

Solve the system of equations:

x-y=2 becomes x=y+2

plug into 4x+4y=32 to solve for y

4(y+2)+4y=32---> 4y+8+4y=32--->8y=24---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 1 Answer:

Equivalent system: 4x+4y=32, x-y=2; solution: x=5, y=3

Problem 2:

Keep Equation 1 the same. Add 1 and 2.

To add an equation, add the left sides together, and then the rights.

so: x+y=8 + x-y=2 gives us: 2x=10

solve for x ---> 2x/2=10/2--->x=5

plug x into x+y=8--->5+y=8--->y=3

Problem 2 answer:

Equivalent system: x+y=8, 2x=10; solution:x=5, y=3

Problem 3:

Subtract Equation 2 from 1, and keep 2 the same.

To subtract an equation, subtract the left sides, then the rights. We are subtracting 1<em> from </em>2, so its 2-1.

x-y=2 - x+y=8 gives us: -2y=-6

Solve for y by dividing by -2-->-2y/-2=-6/-2---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 3 answer:

Equivalent system: -2y=-6, x-y=2; solution: x=5, y=3

Problem 4:

Multiply the sum of Equation 1 and 2 by a factor of 3. Keep equation 2 the same.

First we add 1 and 2: (we did this earlier) ---> 2x=10 ---> now we multiply it all by 3---> 2x*(3)=10*(3)---> this gives us: 6x=30---> now divide by 6 to solve for x: 6x/6=30/6 gives us: x=5

Now, solve for y by plugging x into equation 2: x-y=2---> 5-y=2--->y=3

Problem 4 answer:

Equivalent system: 6x=30, x-y=2; solution: x=5, y=3

______

Quick Tip: One thing inherent of Equivalent systems is that they have the same set of solutions. Thus, we know the systems are equivalent when they have the same set of solutions for x and y. Moreover, you don't need to solve every time after you attempt to find an equivalent system, instead, just plug in the values found in problem 1 to each new set of equations to test if they are equivalent.

If we find x=5 and y=3 for x+y=8 and x-y=2, then all we have to do is plug them in to 6x=30 and -2y=-6 to see if they are equivalent.

6(5)=30 ---> true

-2(3)=-6 ---> true

8 0
4 years ago
force of 400 N stretches a spring 2 m. A mass of 50 kg is attached to the end of the spring and is initially released from the e
andreev551 [17]

Answer:

x = -5 sin (2t)

Step-by-step explanation:

k is the spring stiffness.  The unstretched length of the spring is L.

When the mass is added, the spring stretches to an equilibrium position of L+s, where mg = ks.  When the mass is displaced a distance x (where x is positive if the displacement is down and negative if it's up), the spring is stretched a total distance s + x.

There are two forces on the mass: weight and force from the spring.  Sum of the forces in the downward direction:

∑F = ma

mg − k(s + x) = ma

mg − ks − kx = ma

Since mg = ks:

-kx = ma

Acceleration is second derivative of position, so:

-kx = m d²x/dt²

Let's find k:

F = kx

400 = 2k

k = 200

We know that m = 50.  Substituting:

-200x = 50 d²x/dt²

-4x = d²x/dt²

d²x/dt² + 4x = 0

This is a linear second order differential equation of the form:

x" + ω² x = 0

The solution to this is:

x = A cos (ωt) + B sin (ωt)

Here, ω² = 4, so ω = 2.

x = A cos (2t) + B sin (2t)

We're given initial conditions that x(0) = 0 and x'(0) = -10 (remember that down is positive and up is negative).

Finding x'(t):

x' = -2A sin (2t) + 2B cos (2t)

Plugging in the initial conditions:

0 = A

-10 = 2B

Therefore:

x = -5 sin (2t)

7 0
4 years ago
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