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xz_007 [3.2K]
3 years ago
9

Solve each multiplication problem, use the cancellation method if possible, and reduce your answers.

Mathematics
2 answers:
Alexandra [31]3 years ago
6 0
A. 3/8
b. 8/105
c. 1/4
d. 24
e. 64
f. 29
g. 27 1/3
h. 14 2/3
Mariana [72]3 years ago
4 0
You should google the worksheet that this is from, easier results.

You might be interested in
there are 28 books on the shelf.one-fourth of the books are about animals. two are adventure. the rest are mystery.How many book
yanalaym [24]

Answer:

19

Step-by-step explanation:

1/4 are animal. 28/4 = 7

2 are adventure.

7+2 = 9

9 are not mystery

28-9 = 19

19 are mystery.

4 0
3 years ago
What’s the value of x for a-d?
patriot [66]

Answer:

a = 76 b = 22 c = 58 d = 68

Step-by-step explanation:

A. 180 - 133 = 47

so you have two angles found, which is 57 and 47. Add them together and subtract from 180. You get 76

B. 68 + 90 = 158

You subtract the 158 from 180 and get 22

C. This took some guess work, but when you multiply 58 by 2 you get 116. When subtracted from 180 you get 64. If you add 58 + 58 + 64, you get 180, so this made sense

D. There's an actual rule in trig that explains this, but I forgot it. If you look at the top angle, the other side of the line added to the 68 is a right angle. So you subtract 68 from 90 and get 12. Because there is already the right angle in the triangle, you just need the other two angles to equal 90, and because of the 68 that we found in the other triangle, this made sense.

Sorry, not the best at explanations. Hope that helped

8 0
2 years ago
Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
oksano4ka [1.4K]

Answer: -1.0018*10^42

Step-by-step explanation:

The Euler Method in numerical analysis is used to approximate the solution to an initial value problem using the tangent line to the solution curve through the (x0,y0) to obtain such approximations.

The euler method equation is:

Yn+1 = Yn + h*f(Xn,Yn)

Where n = number of steps, h = Xn+1 - Xn, f(Xn,Yn) is the slope of the curve at (Xn,Yn).

Variables given in the equation

Y(X=0) = 9,

X0 = 0.

h = step = 0.2

f(X,Y) = X^2*Y - 12*Y^2

B

For the first step, n = 0 in the euler equation. Therefore we have:

Y1 = Y0 + h*f(X0,Y0)

Substituting the Y0 = 9 and X0 = 0 into the function f(X,Y) = X^2*Y - 12*Y^2 = 0^2*9 - 12*9^2 = -972.

Therefore Y1 = 9 + 0.2*(-972) = -185.4.

For n = 1

Repeating the same step with X1 = X0 + h = 0 + 0.2 = 0.2,

Y1 = -185.4,

h = 0.2.

Substitute the variables into the equation Y2 = Y1 + h*f(X1,Y1) and f(X1,Y1) = X1^2*Y1 - 12*Y1^2

Y2 = -82682.4672.

Continue the iterations following the steps above till the result is reached.

The summary of the iteration.

When n = 0, X = 0, Y = -185.4

When n = 1 , X = 0.2, Y = -82682.4672

When n = 2 , X = 0.4, Y = -1.6407*10^10

When n = 3, X = 0.6, Y = -6.4609*10^20

When n = 4, X = 0.8, Y = -1.0018*10^42

4 0
2 years ago
Can someone help me with this question
maria [59]

Step-by-step explanation:

1/5 + 3/5 = 3/5 + 1/5 = 4/5

I am not sure what your teacher wants to see on the right side of the second equation.

it would be the whole equation again :

4/5 = 1/5 + 3/5 = 3/5 + 1/5

7 0
2 years ago
Please help! I cannot figure this out, I have no idea what to do.
joja [24]

3. First you should identify which line corresponds to which equation.

x + 2y = -8   ⇒   y = -x/2 - 4

is the line with slope -1/2 and intercept (0, -4), so the first inequality refers to the lower line in the graph.

For the inequality itself, the solution set that satisfies it is either the region above or below it. To decide which, pick any point in the plane and plug its coordinates into the inequality. The origin is a natural choice, since it's not on the line and working with 0s is easy.

In the first inequality, we have

x = 0 and y = 0   ⇒   0 + 2•0 = 0 < -8

which is not true. So (0, 0) is not in the solution set, and it stands to reason that any point in the same region will not be a solution. In other words, the region above the line x + 2y = -8 does not satisfy the inequality, while the one below it does.

In the second inequality,

x = 0 and y = 0   ⇒   2•0 + 0 = 0 ≥ -1

which is true. This means the region containing (0, 0) above the upper line solves the second inequality.

The solution to the system of inequalities is then the intersection of these two regions. (See attached; I've labeled the solution to the first inequality in blue, the solution to the second one in red, and the solution to both in green.)

All this work is kind of overkill for this particular question, though. All you really need to do is check if the given point satisfies the inequalities:

• (-5, 5) :

x = -5 and y = 5   ⇒   -5 + 2•5 = 5 < -8   ⇒   NO

• (2, -5) :

x = 2 and y = -5   ⇒   -2 + 2•5 = 8 < -8   ⇒   NO

• (-4, -6) :

x = -4 and y = -6   ⇒   -4 + 2•(-6) = -16 < -8   ⇒   MAYBE

x = -4 and y = -6   ⇒   2•(-4) + (-6) = -14 < -8   ⇒   YES

• (5, -7) :

x = 5 and y = -7   ⇒   5 + 2•(-7) = -9 < -8   ⇒   MAYBE

x = 5 and y = -7   ⇒   2•5 + (-7) = 3 < -8   ⇒   NO

4. x is the number of candy boxes with chocolates and y is the number of boxes without chocolates. Each of the x boxes cost $27.50, so if you have x boxes, their total cost is $27.5x. Each of the y boxes cost $25.00, so y boxes cost $25y. The sponsor doesn't want to order more than 100 boxes total, so

x + y < 100

but wants to raise at least $2000, so

27.5x + 25y ≤ 2000

Now do the same thing as before; plug in the listed x- and y-coordinates and pick the point that satisfies both inequalities. You will find that (50, 30) is the correct choice.

5. Consult the "overkill" part of problem 3. The line y = -3x + 6 has a negative slope, so it's the downward sloping one. Check if the origin satisfies the inequality:

x = 0 and y = 0   ⇒   0 ≤ -3•0 + 6   ⇒   0 ≤ 6   ⇒   YES

This means Regions II and III solve the first inequality.

Do the same with the other inequality:

x = 0 and y = 0   ⇒   0 ≥ 1/2•0 + 1   ⇒   0 ≥ 1   ⇒   NO

This tells us that Regions I and II solve the second inequality.

The intersection of these regions is of course Region II.

6. One of the plotted lines is apparently y = x + 4, which has a positive slope, so this must be the upper line. The shaded region below it corresponds to the solution of either y < x + 4 or y ≤ x + 4 and we eliminate B.

The other line has negative slope, which eliminates A and D since

3x - 4y = 20   ⇒   y = 3/4x - 5

has positive slope. This leaves D.

6 0
2 years ago
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