Which expresion has a value grater than 4 3 ?

A Ferrari drove 300 kilometers in one and a half hours. How fast was the car going?

irga5000 [103]

I'm assuming you mean how many mph when you refer to how fast the car was going.

So 300 km = 1.5 hrs

1.) Find the number of km per hour
Since we have the equation 300km = 1.5hr, we can just divide 1.5 (to isolate the number of hours) on both sides.

You should get 200 km = 1 hour

2.) Convert km to miles
To do this, you must know the conversion 1 km = 0.621371192 miles

So to convert 200 km to miles, just multiply 200 x 0.621371192.

You should get 124.274238.

That means the Ferrari drove at a pace of roughly 124 mph :)

8 0

3 years ago

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There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con

grandymaker [24]

Answer:

SAS theorem

Step-by-step explanation:

Given

$\square ABCD$

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$

$\angle BAC = \angle DCA$

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

$\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$

and

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$

So, the congruent sides and angles of △ABE and △CDE are:

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$ ---- S

$\angle BAC = \angle DCA$ ---- A

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$ or $\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$ --- S

Hence, the theorem that compares both triangles is the SAS theorem

4 0

2 years ago

Sari applied the distributive property using the greatest common factor to determine the expression that is equivalent to 84 + 4

valentina_108 [34]

Sari did not use the greatest common factor in the equation.

6 0

3 years ago

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Help please !!!!!!!!!!!!!!

ohaa [14]

To start, we're given the range that x lies in: from -1 to 4. We know from the fact that $-1 \leq x$ that -1 will be included in that range, so we mark -1 on the number line with a solid circle. We also know from $x\ \textless \ 4$ that, while x can be any value up to 4, it does not include 4. We indicate this by drawing a hollow circle around 4 on the number line. Since x can be any value within this range, we make that fact clear by drawing a bold line between the points -1 and 4 on the number line. I've attached an image of what our final graph would look like.

5 0

2 years ago

Please help me with the following questions. Thanks!

Natali5045456 [20]

Answer:

The answer to your questions are in bold

Step-by-step explanation:

a)

C = $\left[\begin{array}{ccc}-6&6\\-2&4\\\end{array}\right]$

= -24 + 12

= -12

b) -1 7 -4 -1 -1 - 4 7 - 1 - 5 6

-2 -6 -8 8 = -2 - 8 -6 + 8 -10 2

2 -3 2 -7 2 + 2 -3 - 7 4 -10

-1 10 -6 5 -1 - 6 10 + 5 -7 15

5 0

3 years ago