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Romashka [77]
3 years ago
7

–10 = –6 + 2c –8 1 –2 2

Mathematics
1 answer:
guapka [62]3 years ago
6 0

–10 = –6 + 2c

+ 6     + 6          Add 6 to cancel out -6

-4 = 2c

/2    /2              Divide by 2 to get c variable alone

-2 = c               Choice C is the correct answer

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What type of solutions will a quadratic equation have when the discrimination b^2 - 4ac in the quadratic formula is negative? (p
lys-0071 [83]

Answer:

Step-by-step explanation:

The discriminant comes from the quadratic formula. We use the discriminant to determine, before factoring, what types of solutions we can expect when we do factor. In the quadratic formula, the discriminant is under a square root sign. If the number under a square root sign is negative (meaning the discriminant is negative), we will not have real number solutions because you can't have a negative number under a square root. So a is your choice.

5 0
3 years ago
Find the length of the bold arc. Round to the nearest tenth and use 3.14 or the pi button on your calculator for pi.
Elanso [62]

Answer:

8π yards

Step-by-step explanation:

A circle subtends a total angle of 360 ° from its center.The length of an arc is directly proportional to the angle it subtends from the circle's center. The arc's length can therefore be calculated as:

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4 0
4 years ago
Haydn made a scale model of a building.The actual height of the building is 75 feet. Haydn's model used a scale in which 1.5 inc
elixir [45]

Answer:

3.75 inches

Step-by-step explanation:

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75/30=2.5

1.5 * 2.5=3.75

hope that makes sence. :)

8 0
3 years ago
Read 2 more answers
Find the diagonal of a square whose sides are of the given measure. Given = 5"
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Use the Pythagorean Theorem.  a = b = 5 so
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5 0
4 years ago
Read 2 more answers
Write down a differential equation of the form dy/dt = ay + b whose solutions have the required behavior as t → ∞. all other sol
nordsb [41]

The ODE has an equilibrium point at

\dfrac{\mathrm dy}{\mathrm dt}=ay+b=0\implies y=-\dfrac ba

We want y=2 to be an unstable equilibrium solution - in other words, we want all solutions with initial condition y(t_0)=c for c near 2 to diverge from y=2 - so -dfrac ba=2.

Divergence from y=2 would require that for y>2, any solution is increasing, and for y, and solution is decreasing. In order for that to happen, we need

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\dfrac{\mathrm dy}{\mathrm dt}=ay+b

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Now we just pick a,b to satisfy these conditions. An easy choice is a=1, which forces b=-2, so that one possible ODE would be

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I've attached a plot of the slope field to demonstrate the behavior of the solutions.

6 0
3 years ago
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