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oee [108]
3 years ago
12

Evaluate the expression when b=3 and c=-6 4b-c

Mathematics
1 answer:
sveta [45]3 years ago
6 0

Answer:

18

Step-by-step explanation:

4(3) -(-6)

4 times 3 is 12. When you subtract a negative number the negative number gets added so it would be 12 + 6 which is 18. Hope that helps :)

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A sled is being pulled across a floor by two ropes such that the angle between them is 40°. If the forces on the ropes are 100 p
Dafna1 [17]

Answer:

option 4 ⇒ 236 lb.

Step-by-step explanation:

Best explanation of the question is as shown in the attached figure.

we will use the parallelogram method to calculate resultant force.

to get the length of the resultant force ⇒ use the cosines law

The cosine law is a² = b² + c² - 2 * b * c * cos (∠A)

Applying at the question where b = F₁  , c = F₂  and  ∠A = ∠x

Given that F₁ = 100 pounds  , F₂ = 150 pounds  and  ∠x = 180° - 40° = 140°

∴ (Resultant force)² = 100² + 150² - 2 * 100 * 150 * cos (∠140) = 55481

∴ Resultant force = √55481 = 235.54 ≅ 236 pounds

The answer is option 4 ⇒ 236 lb.

8 0
3 years ago
Quadratic inequalities
lina2011 [118]
See picture for rough graph

7 0
3 years ago
Hello can someone help me with algebra 1 homework ?
lbvjy [14]
Whats the problem? I could help lol
3 0
3 years ago
(a) By inspection, find a particular solution of y'' + 2y = 14. yp(x) = (b) By inspection, find a particular solution of y'' + 2
SOVA2 [1]

Answer:

(a) The particular solution, y_p is 7

(b) y_p is -4x

(c) y_p is -4x + 7

(d) y_p is 8x + (7/2)

Step-by-step explanation:

To find a particular solution to a differential equation by inspection - is to assume a trial function that looks like the nonhomogeneous part of the differential equation.

(a) Given y'' + 2y = 14.

Because the nonhomogeneus part of the differential equation, 14 is a constant, our trial function will be a constant too.

Let A be our trial function:

We need our trial differential equation y''_p + 2y_p = 14

Now, we differentiate y_p = A twice, to obtain y'_p and y''_p that will be substituted into the differential equation.

y'_p = 0

y''_p = 0

Substitution into the trial differential equation, we have.

0 + 2A = 14

A = 6/2 = 7

Therefore, the particular solution, y_p = A is 7

(b) y'' + 2y = −8x

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x

2Ax + 2B = -8x

By inspection,

2B = 0 => B = 0

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x

(c) y'' + 2y = −8x + 14

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x + 14

2Ax + 2B = -8x + 14

By inspection,

2B = 14 => B = 14/2 = 7

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x + 7

(d) Find a particular solution of y'' + 2y = 16x + 7

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = 16x + 7

2Ax + 2B = 16x + 7

By inspection,

2B = 7 => B = 7/2

2A = 16 => A = 16/2 = 8

The particular solution y_p = Ax + B

is 8x + (7/2)

8 0
3 years ago
What is the math problem answer?
azamat

To find the mid point add both X coordinates together and divide by 2 and then do the same with the y coordinates.

X coordinate: 4 + -6 = 4-6 = -2 / 2 = -1

Y coordinate: -2 + 3 = 1 / 2 = 1/2


Midpoint: ( -1, 1/2)

3 0
3 years ago
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