All categories

3 years ago

Look at the graph of this system of equations: y = - x2 + 1 and y = x2. At which approximate points are the two equations equal?

Mathematics

2 answers:

Alexandra [31]3 years ago

8 0

Y=-x^2+1 and y=x^2

When the two curves intersect their coordinates will be equal, so we can say y=y whenever a solution exists, so:

x^2=-x^2+1 add x^2 to both sides

2x^2=1 divide both sides by 2

x^2=1/2 take the square root of both sides

x=±√(1/2), so there are two solutions, we can use x^2 to find the corresponding y values.

y=x^2, y=1/2 in each instance, so the two point where the curves intersect are:

(-√(1/2), 1/2) and (√(1/2), 1/2) or if you want approximations....

(-0.7, 0.5) and (0.7, 0.5)

NemiM [27]3 years ago

6 0

Answer: they are correct....

Step-by-step explanation: I know this cause I just took the mastery test and got it right.

You might be interested in

Ron made 3 pitchers of lemonade to sell at a lemonade stand. The cost of making the lemonade is 50 cents per pitcher of lemonade

julia-pushkina [17]

Answer:

3.90 I think.

Step-by-step explanation:

first, 15×36. then subtract 50 three times.

6 0

3 years ago

Read 2 more answers

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

Can someone help me with this? I don’t need an explanation I just need the answer thank you :)

Fittoniya [83]

Answer:

87.92 in²

Step-by-step explanation:

pi × (R² - r²) = area

Where R is the radius of the outer circle and r of the inner circle.

Since C = 2pi×r

37.68 = 2×3.14×r

r = 6 in

R = 6 + 2 = 8 in

Therefore,

pi × (R² - r²) = area

Area = 3.14 (8² - 6²)

= 87.92 in²

5 0

3 years ago

What’s the answer to this equation <br> 3x+6t=?

GaryK [48]

are there any options

7 0

3 years ago

What IS the simplified value of the expression below? -2.6 + 5.8

Lerok [7]

Answer:

3.2

Step-by-step explanation:

5 0

3 years ago