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lubasha [3.4K]
3 years ago
12

70 POINTS

Mathematics
2 answers:
mamaluj [8]3 years ago
3 0

Answer: A

Step-by-step explanation:

nikitadnepr [17]3 years ago
3 0

Answer:

d

Step-by-step explanation:

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What is the first quartile of the set of data shown below?
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The answer would be B 23
7 0
3 years ago
2. Two angles have a common vertex but no common side. How can you tell whether the angles are vertical angles?
Maksim231197 [3]

Explanation:

The angles are <em>vertical angles</em> if the opposites of the rays forming one of the angles are the rays forming the other angle.

More formally, if V is the common vertex, and ...

  • R is a point on one of the rays forming Angle 1
  • S is a point on the ray that is the opposite of ray VR
  • T is a point on the other ray forming Angle 1
  • U is a point on the ray that is the opposite of ray VT

Then angle RVT and angle SVU are vertical angles.

___

Another way to say this is that points R, V, S are collinear, as are points T, V, U, and the two angles of interest are RVT and SVU.

If the above conditions cannot be met, then the angles are not vertical angles.

7 0
3 years ago
Julian is using a biking app that compares his position to a simulated biker traveling Julian's target speed. When Julian is beh
zepelin [54]

Answer:

\large \boxed{\text{11 km/h}}

Step-by-step explanation:

1. The situation after 15 min:

(a) Distance travelled by simulated biker:

15 min =¼ h  

\text{Distance} = \dfrac{1}{4}\text{ h} \times \dfrac{\text{20 km}}{\text{1 h}} = \text{5 km}

(b) Distance travelled by Julian

Julian is 2¼ km behind  the biker. The distance he has travelled is (5 - 2¼) km

5 - 2¼ = 5 - ⁹/₄ = ²⁰/₄ - ⁹/₄ = ¹¹/₄ = 2¾

Julian has travelled 2¾ km in ¼ h.

2. Julian's average speed

\text{Speed} = \dfrac{\text{Distance}}{\text{Time}} = \dfrac{2\frac{3}{4}\text{ km}}{\frac{1}{4}\text{ h}} =\dfrac{11}{4}\text{ km}\times \dfrac{4}{\text{1 h}} = \textbf{11 km/h}\\\\\text{Julian's average speed is $\large \boxed{\textbf{11 km/h}}$}

7 0
3 years ago
Read 2 more answers
Find an equation of the plane through the point (−1,−5,−2)(−1,−5,−2) with normal vector n=⟨−5,−4,−1⟩n=⟨−5,−4,−1⟩.
Ostrovityanka [42]
Consider this option:
1. given: n(A;B;C)=n(-5;-4;-1) and A(-1;-5;-2)=A(x₀;y₀;z₀).
Common view of equation for a plane is Ax+By+Cz+D=0, where A,B,C,D - numbers.
2. from another side using coordinates of A and normal vector it is possible to make up the equation: A(x-x₀)+B(y-y₀)+C(z-z₀)=0 ⇒ -5(x+1)-4(y+5)-(z+2)=0; ⇒ -5x-4y-z-27=0 or 5x+4y+z+27=0.
7 0
3 years ago
Convert each function to standard form<br>31) y=(x + 2)2 + 1​
vlada-n [284]

The standard form of a parabola is

y=ax^2+bx+c

So, if you expand the square, you get

y=(x+2)^2+1 = x^2+4x+4+1 = x^2+4x+5

which is the standard form

6 0
3 years ago
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