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3 years ago

1. Work out the speed of a car travelling:

Mathematics

2 answers:

Norma-Jean [14]3 years ago

7 0

Answer:

10m/s,15m/s,20m/s,5m/s

Step-by-step explanation:

Speed = distance / time

a) distance = 100m , time = 10s

Speed = distance / time

Speed = 100m/10s

Speed = 10m/s

b) distance = 300m , time = 20s

Speed = distance / time

Speed = 300m/20s

Speed = 15m/s

c) distance = 700m , time = 35s

Speed = distance / time

Speed = 700m/35s

Speed = 20m/s

d) distance = 1000m , time = 200s

Speed = distance / time

Speed = 1000m/200s

Speed = 5m/s

I hope this was helpful, please rate as brainliest

egoroff_w [7]3 years ago

6 0

Step-by-step explanation:

speed is a scalar quantity (which only needs a magnitude), and velocity is a vector quantity (needs a magnitude with a direction)

Here since they want the "speed" you can use the velocity formula, but without a direction.

$velocity = \frac{distance}{time}$

A. v=100/10

v=10m/s

B. v=300/20

v=15m/s

C. v=700/35

v=20m/s

D. V=1000/200

v=5m/s

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Answer:

The final amount is $1109.81

Step-by-step explanation:

In order to find the total amount, start with the know amount, which is Ms. Moore's class. Her class raised $249. Now we can use that to find the amount from Ms. Aguilar's class.

$249 + $396.62 = $645.62

Now we can use the amount from Ms. Aguilar's class to find the amount from Ms. Barry's class

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2 years ago

Two angles of a quadrilateral measure 200° and 50°. The other two angles are in a ratio of 2:9. What are the measures of those t

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Answer:

20° and 90°

Step-by-step explanation:

Let 2x = measure of 1st angle

then 9x = measure of 2nd angle

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2 years ago

In the adjoining figure, XY = XZ . YQ and ZP are the bisectors of <img src="https://tex.z-dn.net/?f=%20%5Cangle" id="TexFormula1

svetlana [45]

Answer:

See Below.

Step-by-step explanation:

Statements: Reasons:

$1)\, XY=XZ$ Given

$2) \text{ $ m\angle Y= m\angle Z$}$ Isosceles Triangle Theorem

$\displaystyle 3) \text{ $m\angle Y=m\angle XYQ + \angle QYZ$}$ Angle Addition

$\displaystyle 4)\text{ $YQ$ bisects $\angle XYZ$}$ Given

$5) \text{ $m\angle XYQ=m\angle QYZ$}$ Definition of Bisector

$\displaystyle 6)\text{ $m\angle Y=2m\angle QYZ$}$ Substitution

$7)\text{ $m\angle Z=m\angle XZP+m\angle PZY$}$ Angle Addition

$8)\text{ $ZP$ bisects $\angle XZY$}$ Given

$\displaystyle 9) \text{ $m\angle XZP=m\angle PZY$ }$ Definition of Bisector

$\displaystyle 10) \text{ $ m\angle Z = 2m\angle PZY $}$ Substitution

$11)\text{ } 2m\angle QYZ=2m\angle PZY$ Substitution

$12)\text{ }m\angle QYZ=m\angle PZY$ Division Property of Equality

$13)\text{ } YZ=YZ$ Reflexive Property

$14)\text{ } \Delta YZP\cong\Delta ZYQ$ Angle-Side-Angle Congruence*

$15)\text{ } YQ=ZP$ CPCTC

*For clarification:

∠Y = ∠Z

YZ = YZ (or ZY)

∠PZY = ∠QYZ

So, Angle-Side-Angle Congruence:

ΔYZP is congruent to ΔZYQ

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