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zheka24 [161]
3 years ago
12

Someone help pls. can someone solve it for me??

Mathematics
1 answer:
cupoosta [38]3 years ago
3 0

Answer:

  slope = 5/3

Step-by-step explanation:

The line rises 5 units for each 3 units it goes to the right. (The y- and x-intercepts are clues.) The slope is the ratio of rise to run: 5/3.

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Solve for x in the inequality.
oee [108]

Answer:

C. x < 5

Step-by-step explanation:

12x < 60

x < 5

5 0
3 years ago
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What is the measure of XYZ​
lorasvet [3.4K]

Answer:

B

Step-by-step explanation:

The arc XZ is twice the measure of the inscribed angle XYZ , that is

arc XZ = 2 × 60° = 120°

The complete circumference = 360° , thus

arc XYZ = 360° - arc XZ = 360° - 120° = 240° → B

6 0
2 years ago
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Here is an equilateral triangle. The length of each side is units. A height is drawn. In an equilateral triangle, a line drawn f
Temka [501]

1. Height of the equilateral triangle is: √3 units

2. Area of the equilateral triangle = √3 units²

3. Area = √3/4 x², when each side is x.

<h3>What is an Equilateral Triangle?</h3>

A triangle that has all its three sides equal in length, is referred to as an equilateral triangle.

1. Given the each side measures 2 units, and h is the height, applying the Pythagorean theorem, we would have:

h = √(2² - 1²)

h = √(4 - 1)

h = √3

2. Area of the equilateral triangle = 1/2bh = 1/2(2)(√3)

Area of the equilateral triangle = √3 units²

3. If x is the side length of the equilateral triangle, we would have:

height (h) = √[x² - (x/2)²] = √[x² - x²/4] = √(3x²/4) = √3/2x

Area = 1/2bh = 1/2(x)(√3/2x) = √3/4 x²

Learn more about equilateral triangle on:

brainly.com/question/15294703

#SPJ1

3 0
1 year ago
I need to solve: 6x+3^59
MrMuchimi
6х=59:3
6х=19,6
Х=19,6:6
Х=3,2
7 0
3 years ago
A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th
olchik [2.2K]

Answer:

14.52 seconds.

Step-by-step explanation:

We have been given that the height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation y=-16x^2+224x+121. We are asked to find the time, when the rocket will hit the ground.

We know that the rocket will hit the ground, when height will be 0. So to find the time when rocket will hit the ground, we will substitute y=0 in our given equation as:

0=-16x^2+224x+121

Let us solve for x using quadratic formula.

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\frac{-224\pm\sqrt{224^2-4(-16)(121)}}{2(-16)}

x=\frac{-224\pm\sqrt{50176+7744}}{-32}

x=\frac{-224\pm\sqrt{57920}}{-32}

x=\frac{-224\pm240.66574}{-32}

x=\frac{-224+240.66574}{-32}, x=\frac{-224-240.66574}{-32}

x=\frac{16.66574}{-32}, x=\frac{-464.66574}{-32}

x=-0.520804375, x=14.520804375

Upon rounding to nearest 100th of second, we will get:

x\approx -0.52, x\approx 14.52

Since time cannot be negative, therefore, the rocket will hit the ground after 14.52 seconds.

6 0
3 years ago
Read 2 more answers
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