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monitta
3 years ago
6

Henry draws an obtuse triangle how many obtuse angles does Henry's triangle have? Please eye help

Mathematics
1 answer:
Travka [436]3 years ago
8 0
The sum of all three angles in any triangle is always 180 degrees. So in obtuse triangle there is only one obtuse angle (more than 90 degrees). If we have for example an obtuse angle of 91 degrees, then two other angles have to fit in 89 degrees which makes it impossible for any of them to be obtuse.
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Solve for p -10p = -40 ???
scoundrel [369]

Answer:

p = 40/9

Step-by-step explanation:

p - 10p = -40

-9p = -40

p = -40/(-9

p = 40/9

8 0
3 years ago
Let △ABC be a right triangle with m∠C = 90°. Given tan ∠B = 1, find tan ∠A.
patriot [66]
A triangle is equaled to 180 degrees
Si you know your B and C do all you do is
180-90-1 =
89
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4 0
3 years ago
Find the equation of a line that contains the points (−3,−1) and (−4,−7). Write the equation in slope-intercept form
kipiarov [429]

Answer:

is the answer

11 upon 4 (11/4)

6 0
3 years ago
A rectangular vegetable patch has a perimeter of 28 meters and an area of 48 square
aliina [53]

Answer:

Length = 8 meters

Width = 6 meters

Step-by-step explanation:

Given the following information:

Perimeter of a rectangular vegetable patch: 2(L + W) = 28 meters

Area = L × W = 48 m²

We can solve for the dimensions by using both equations.

First, let's use the formula for the perimeter of the rectangular vegetable patch, and isolate one of the given variables:

2(L + W) = 28

Divide both sides by 2:

\frac{2(L + W)}{2} = \frac{28}{2}

L + W = 14

Subtract W from both sides to isolate L:

L + W - W = 14 - W

L = 14 - W

Next, we'll take the formula for the area, and substitute the value for the L from our previous step:

A = 48  =  L × W

48  = W × (14 - W)

Distribute W into the parenthesis:

48 = 14W - W²

Add W² and subtract 14W to both sides:

W² - 14W + 48 = 14W - 14W - W² + W²

W² - 14W + 48 = 0   ⇒ This represents a quadratic equation in standard form. We can use the coefficient and constant values to solve for its roots.

a = 1, b = -14, and c = 48

Substitute these values into the quadratic equation:

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

x = \frac{14 +/- \sqrt{(-14)^{2}-4(1)(48)}}{2(1)}

x = \frac{14 +/- \sqrt{196-192}}{2}

x = \frac{14 +/- \sqrt{4}}{2}

x = \frac{14 + 2}{2}, x = \frac{14 - 2}{2}

x = 8, x = 6

Now, we can substitute these values into the formulas for the perimeter and area to find the true dimensions of the rectangular vegetable patch.

Perimeter: 2(L + W) = 2(8 + 6) = 28 meters

Area = L × W = 8 × 6  = 48 m²

Therefore, the dimensions of the rectangular vegetable patch are:

Length = 8 meters

Width = 6 meters

Dimensions: 8 × 6 meters.

3 0
3 years ago
Solve for x and y x+y=15 x+2y=23
BigorU [14]

Answer:

x=7, y= 8

Step-by-step explanation:

x+y=15

x+2y=23

Subtract the first equation from the second equation to eliminate x

x+2y=23

-x-y=-15

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

  y = 8

Now we can find x

x+y = 15

x+8 = 15

Subtract 8 from each side

x+8-8=15-8

x = 7

5 0
3 years ago
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