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nika2105 [10]
3 years ago
14

10 Determine the area of the figure. 9 cm 6 cm 12 cm​

Mathematics
1 answer:
Valentin [98]3 years ago
8 0

Answer:

Area= 648

Step-by-step explanation:

Area formula: L times W times H

9 times 6 times 12        :)

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A bike trail is drawn on a coordinate system. Each end of the trail is an x-intercept. The height of the trail is modeled by the
Aleks [24]

Answer:

4metres

Step-by-step explanation:

Given the height of the trail modeled by the function h(x) = -0.1x^2 - 0.3x +2.8 , where x is the horizontal distance in kilometers

The horizontal distance occurs at the x intercept i.e at h(x) = 0

Substitute

0 = -0.1x^2 - 0.3x +2.8 ,

Multiply through by -10

0 = x^2 + 3x - 28

x^2 + 3x - 28 = 0

Factorize

x^2 + 7x - 4x - 28 = 0

x(x+7) - 4 (x+7) = 0

(x-4)(x+7) = 0

x-4 = 0 and x+7 = 0

x = 4 and -7

Since the danced cannot be negative, hence the horizontal distance of the trail is 4m

5 0
3 years ago
Find the value of X for the triangle.
user100 [1]

Answer:

8

Step-by-step explanation:

77+50=127

180-127=53

7x-3=53

7x=53+3

7x=56

x=56/7

x=8

8 0
3 years ago
Help it for my final
Ghella [55]

Answer:  I THINK ITS A

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
One angle of a right-angled triangle is 60 grades . Find the other angle in degree.​
eduard

Answer:

<u>30°</u>

Step-by-step explanation:

This right triangle has :

  1. right angle (90°)
  2. One angle of 60°

=========================================================

Solving :

We know the angle sum of a triangle is 180°.

⇒ x + 90° + 60° = 180°

⇒ x + 150° = 180°

⇒ x = <u>30°</u>

3 0
2 years ago
Read 2 more answers
1. While solutions to the general quintic equation in terms of radicals are impossible to obtain, they do exist. Prove that the
Marta_Voda [28]

Answer:

Step-by-step explanation:

Given is a function as x^5-3x+1

Equating to 0 we have equation

If the function f(x) has x intercepts then the solutions are real

Let us use remainder theorem and change of signs rule

f(0) = 1>0

f(-1) = -1+3+1=3

f(-2) = -32+6+1<0

This implies there is a real root between -1 and -2.

f(1) = -1

Since f(0) and f(1) have different signs, there exists a real root between 0 and 1.

f(2) = 32-6+1>0

Since f(1) and f(2) have different signs there exists a real root between 1 and 2.

Thus there are definitely three real solutions as

one between -1 and 0, one between 0 and 1, and third between 1 and 2.

3 0
4 years ago
Read 2 more answers
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