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kvv77 [185]
3 years ago
6

I need the area of the below shape

Mathematics
2 answers:
sukhopar [10]3 years ago
4 0
The surface area of the triangular prism is 16cm
Ratling [72]3 years ago
3 0
The answer is B.280 

hope this helps

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What is the circumstances of this circle? Use 3.14 for
Papessa [141]

Answer:

Is

Step-by-step explanation:

Is there more information?

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3 years ago
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What is the range of the function f(x) =4x-1 when the domain is -3 ,0,4 ?
dedylja [7]

Answer:

The range of this function is {-13, -1, 15}.

Step-by-step explanation:

Evaluate f(x) = 4x - 1 at {-3, 0, 4}:

f(-3) = -13

f(0) = -1

f(4) = 15

The range of this function is {-13, -1, 15}.

3 0
3 years ago
Carla had 8 sticks of gum. She gave 2/8 of her gum to her friends. How many sticks of gum does Carla have left?
svetlana [45]


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8 0
4 years ago
Solving systems of linear inequalities
PIT_PIT [208]

Answer:

YES. (2, 7) is a solution of the system.

Step-by-step explanation:

System of linear inequalities has been given as,

y ≥ -x + 1 --------(1)

y < 4x + 2 ------(2)

If (2, 7) is a solution of the given system of inequalities, it will satisfy both the inequalities.

By substituting the coordinates of point (2, 7) in inequality (1),

7 ≥ -2 + 1

7 ≥ -1

True.

By substituting the coordinates of point (2, 7) in inequality (2),

7 < 4(2) + 1

7 < 9

True.

Therefore, point (2, 7) lie in the solution area of system of inequalities.

YES. (2, 7) is a solution of the system.

4 0
3 years ago
PLEASE HELP QUICK!! WILL OFFER 100 POINTS TO THE FIRST ONE THAT ANSWERS
Reil [10]

Given

x+1 = \sqrt{7x+15}

We have to set the restraint

x+1\geq 0 \iff x \geq -1

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}

So, we have to impose

x-7\geq 0 \iff x \geq 7

Squaring both sides, we have

(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

7 0
3 years ago
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