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Mars2501 [29]
3 years ago
15

Help me please!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Alla [95]3 years ago
6 0

mr. russell questions:

1: false.

2: false.

3: true

4: false.

5: true.

the last question: D.

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andrew-mc [135]
Data:
r (radius) = 14 m
h (height) = 6 m
v (volume) = ?

Formula:
V =  \pi *r^2*h

Solving:
V = \pi *r^2*h
V =  \pi *14^2*6
V =  \pi *196*6
\boxed{\boxed{V = 1176 \pi\:m^3}}\end{array}}\qquad\quad\checkmark
4 0
3 years ago
HELP PLEASE!!!!!!! I NEED FAST!!
forsale [732]

The answer is b) 8.625.

Add green and red peppers, then divide it by 1.5.

5 0
3 years ago
Read 2 more answers
A computer online service charges one hourly rate for regular use but a higher hourly rate for designated "premium" areas. One c
N76 [4]

Answer: the service charge per hour for premium services is $5.5

the service charge per hour for regular services is $3

Step-by-step explanation:

Let x represent the service charge per hour for premium services.

Let y represent the service charge per hour for regular services.

One customer was charged $38 after spending 2 h in premium areas and 9 regular hours. It means that

2x + 9y = 38- - - - - - - - - - - 1

Another customer spent 3 h in premium areas and 6 regular hours and was charged $34.50. It means that

3x + 6y = 34.5- - - - - - - - - - -2

We would eliminate x by multiplying equation 1 by 3 and equation 2 by 2. It becomes

6x + 27y = 114

6x + 12y = 69

Subtracting, it becomes

15y = 45

y = 45/15

y = 3

Substituting y = 3 into equation 1, it becomes

2x + 9 × 3 = 38

2x + 27 = 38

2x = 38 - 27 = 11

x = 11/2 = 5.5

8 0
3 years ago
Explain how you could use the objects to act out this problem
attashe74 [19]
Use some bricks and paint. paint the bricks
4 0
3 years ago
Read 2 more answers
How would you solve these 2 maths problems?
sweet-ann [11.9K]

Answer:

see explanation

Step-by-step explanation:

(9)

Expand and simplify right side and compare coefficients of like terms on left side.

(x - 2)(x² + ax + b) + c

= x³ + ax² + bx - 2x² - 2ax - 2b + c

= x³ + x²(a - 2) + x(b - 2a) - 2b + c

compare with

x³ + 2x² - 3x + 4

x² terms

a - 2 = 2 ( add 2 to both sides )

a = 4

x terms

b - 2a = - 3 ← substitute a = 4

b - 8 = - 3 ( add 8 to both sides )

b = 5

constant terms

- 2b + c = 4 ← substitute b = 5

- 10 + c = 4 ( add 10 to both sides )

c = 14

Thus a = 4, b = 5 and c = 14

-------------------------------------------------

(10)

Since both functions cross the x- axis at - 2 then (- 2, 0) satisfies both, that is

f(- 2) = (-2)^{4} + a(- 2)³ + b(- 2)² + 36(- 2) + 144 = 0, that is

- 16 - 8a + 4b - 72 + 144 = 0

- 8a + 4b + 56 = 0

- 8a + 4b = - 56 → (1)

and

g(- 2) = (-2)^{4} + (a + 3)(- 2)³ - 23(- 2)² + (b + 10)(- 2) + 40 = 0, that is

- 16 - 8(a + 3) - 92 - 2(b + 10) + 40 = 0

- 16 - 8a - 24 - 92 - 2b - 20 + 40 = 0

- 8a - 2b - 112 = 0

- 8a - 2b = 112 → (2)

Subtract (1) from (2) term by term

- 6b = 168 ( divide both sides by - 6 )

b = - 28

Substitute b = - 28 into (2)

- 8a + 56 = 112 ( subtract 56 from both sides )

- 8a = 56 ( divide both sides by - 8 )

a = - 7

Thus a = - 7 and b = - 28

5 0
3 years ago
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