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zysi [14]
4 years ago
7

Ehhh... idk okay?? Help me

Mathematics
1 answer:
Tasya [4]4 years ago
5 0

Answer:

a. 48

b. 2

Step-by-step explanation:

Follow PEMDAS.

PART A

3*(4+5*3-3)

3*(4+15-3)

3*(16)

48

PART B

{10-[5÷(8-3)*2+6]}

{10-[5÷(5)*2+6]}

{10-[1*2+6]}

{10-[2+6]}

{10-[8]}

2

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PLEASE HELP! i have a couple minutes left
swat32

Answer:

1.8

Step-by-step explanation:

slope formula: (y2-y1)/(x2-x1)

(86-32)/(30-0)

54/30

1.8

8 0
4 years ago
Read 2 more answers
1<br> Use f(x) = ½ x and ƒ˜¹(x) = 2x to solve the problems.<br> f(2)=<br> f¹1)=[<br> f(f(2)) =
liq [111]

Answer:

f(2) = 1

(1) = 2

(f(2)) = 2

Step-by-step explanation:

f(x) = x

(x) = 2x

f(2) = (2)

f(2) = 1

(1) = 2(1)

(1) = 2

(f(2)) = 2((2))

(f(2)) = 2(1)

(f(2)) = 2

7 0
2 years ago
H(t)=2t-2 <br> g(t)=t-1 <br> find h(g(2))
Crazy boy [7]
T = 2

To find this, follow these steps.

First, you need to find g(2):

t-1
2-1
1

Therefore 1 is your first answer. Then you input that answer into h(t).

2(t) - 2
2(2) - 2
4 - 2
2

Thus, 2 is your final answer.

Hope this helps!
5 0
3 years ago
Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by h(
andreev551 [17]

Answer:

a. 38.19m/s

b. 38.605m/s

c. 38.937m/s

d. 39.0117m/s

e. 39.01917m/s

Step-by-step explanation:

The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:

v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}

Where:

x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval

a. Let's find h(3) and h(4) using the data provided by the problem:

h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f

The average velocity over the interval [3, 4] is :

v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s

b. Let's find h(3.5) using the data provided by the problem:

h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f

The average velocity over the interval [3, 3.5] is :

v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s

c. Let's find h(3.1) using the data provided by the problem:

h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f

The average velocity over the interval [3, 3.1] is :

v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s

d. Let's find h(3.01) using the data provided by the problem:

h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f

The average velocity over the interval [3, 3.01] is :

v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s

e. Let's find h(3.001) using the data provided by the problem:

h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f

v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s

7 0
3 years ago
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jekas [21]
The correct answer is $3. Please mark me brainiest. :)
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