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Alekssandra [29.7K]
3 years ago
9

Which point is located in quadrant IV?

Mathematics
1 answer:
alex41 [277]3 years ago
6 0
Quadrant General Form of Point in this Quadrant Example
I (+, +) (5, 4)
II (−, +) (−5, 4)
III (−, −) (−5, −4)
IV (+, −) (5, −4)
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PLEASE HELP! 40 POINTS! state the value only do not use units. do not write x= in your response. Thank you
12345 [234]

Answer:

y=43 (alternate angle)

in triangle ADC

97+y+x=180(,sum of interior angle of triangle is 180)

x=180-97-43

x=40

8 0
3 years ago
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A triangle has angles measuring 30 degrees and 90 degrees the length of the included side is 6cm. tell whether the conditions fo
Alexxx [7]
Angle 1 = 30°
Angle 2 = 90°
⇒ Angle 3 = 60°
So, it's right triangle. We can set the length of one side and get all other sides.
So, we have 1 triangle. If there are 2 or more triangles with the same data, all the triangles will be congruent because of : <span>Two triangles are </span><span>congruent if </span>"<span>ASA: Two interior angles and the included side in a triangle have the same measure and length, respectively, as those in the other triangle</span><span>.</span>"
4 0
3 years ago
How many 3/4 foot lengths of pipe can be cut from a 7 and 1/2 foot pipe?
stiks02 [169]

Answer:

The number of pipes to be cut from total length of pipe is 10 pipes

Step-by-step explanation:

Given as :

Total length of pipe = l =  7\frac{1}{2} foot

i.e ' = \frac{15}{2}  foot

The length of pipe to be cut = l' = \dfrac{3}{4} foot

Let The number of pipes to be cut = n pipes

<u>According to question</u>

The number of pipes to be cut = \dfrac{\textrm Total length of pipe}{Length of pipe to be cut}

i.e n = \dfrac{l}{l'}

Or, n = \frac{\frac{15}{2}}{\frac{3}{4}}

Or, n = \frac{15\times 4}{3\times 2}

or, n = 5 × 2

∴ n = 10 pipes

So, The number of pipes to be cut = n = 10 pipes

Hence, The number of pipes to be cut from total length of pipe is 10 pipes Answer

3 0
3 years ago
Eliminate the parameter and obtain the standard form of the rectangular equation. line through (x1, y1) and (x2, y2): x = x1 + t
Helen [10]

The parametric equations for the line passes through the points (X₁ , Y₁) and (X₂ , Y₂) are :

X = X₁ + t (X₂ - X₁) ......................... (1)

Y = Y₁ + t (Y₂ - Y₁) .......................... (2)

For eliminating the parameter (t) , first we will solve any one equation for 't' and then substitute that into another equation.

From the equation (1),

⇒ t = \frac{X- X1}{(X2 - X1)}

Substituting this t = \frac{X- X1}{(X2 - X1)} into the equation (2), we will get :

Y = Y₁ + \frac{(X- X1)}{(X2 - X1)} (Y₂ - Y₁)

Y = Y₁ + \frac{(X- X1)(Y2 - Y1)}{(X2 - X1)}

So, this is the standard form of rectangular equation.

For two given points (0, 0) and (4, -4)

X₁ = 0 , Y₁ = 0, X₂ = 4 and Y₂ = -4

For finding the parametric equations, we will plug these values into the given parametric equations.

X = X₁ + t (X₂ - X₁)

⇒ X = 0+ t (4 - 0)

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and Y = Y₁ + t (Y₂ - Y₁)

⇒ Y = 0+ t (-4 - 0)

⇒ Y = - 4t

So, the parametric equations for the line passing through (0,0) and (4, -4) are: X = 4t and Y = -4t

6 0
3 years ago
So this is wrong ? How do i fix it
Inga [223]

I think the 57 is an whole number.

5 0
3 years ago
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