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AnnZ [28]
2 years ago
6

If both the length and width of a rectangle are doubled, how is the area of the rectangle changed?

Mathematics
1 answer:
Law Incorporation [45]2 years ago
7 0

Answer:

B) The area is four times larger

Step-by-step explanation:

A=LW

2*2*A=2L2W

4A=2L2W

Therefore, the area is four times larger

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I need help ........... here’s a picture of the question
cricket20 [7]

Answer:

The answer is B. m∠ POS + m∠ POR = 180

step-by-step explanation:

Because if you already know that m∠POQ equals 90 because it is a right angle

Then all you have to do is take 90 from 180 which is 90 then divide 90 by two which is 45

now you have the measurement for the two acute angles 45

which makes m∠POS = 45 and m∠QOR =45

NOW YOU HAVE TO ADD 90 AND  45 AND 45 ALL TOGETHER THEN YOU WILL HAVE 180

4 0
4 years ago
If f (x) = 2x +1 and g(x) = x^2 + 2x + 1, find f (g(x)).
Morgarella [4.7K]

Answer:

2x² + 4x + 3

Step-by-step explanation:

Given:

f (x) = 2x +1

g(x) = x² + 2x + 1

to find f(g(x)), simply take the x-terms in f(x) = 2x+1 and replace them with g(x) = x² + 2x + 1

Hence,

f(g(x))

= f(x² + 2x + 1)

= 2(x² + 2x + 1) + 1

= 2x² + 4x + 2 +1

= 2x² + 4x + 3

8 0
3 years ago
Yanika is teaching a swim class. She needs to divide the class into teams for a relay race. There is 18 students.
coldgirl [10]
A possible sizes two teams of 9 or 9 teams of 2, 6 teams of 3 or 3 teams of 6

B yes 3 lanes could divide her 18 students by putting 6 people in each lane 
18/3= 6
8 0
3 years ago
4 + 3x = 3х + 5<br> 4 = 5
Inessa [10]

Answer:

4≠5

There is no solution

Step-by-step explanation:

7 0
3 years ago
Solve each equation for 0 x 360 sqrt 2 cosx+1=0 please hurry
jolli1 [7]

Answer:

120\º \text{ and } 240\º

or in radians:

$\frac{2\pi }{3} \text{ and } \frac{4\pi }{3}$

Step-by-step explanation:

From the way you wrote, you want to solve the equation

\sqrt {2 \cos(x)+1}=0 for 0 \leq  x < 360\º, or in radians [0, 2\pi)

\sqrt {2 \cos(x)+1}=0

<u>Square both sides</u>

2 \cos(x) +1= 0

2\cos(x)=-1

$\cos(x)=-\frac{1}{2} $

In the Unit Circle, considering one revolution (interval [0, 2\pi)),

the values where $\cos(x)=-\frac{1}{2} $  are in Quadrant II and III.

Once

$\cos(x)= \frac{1}{2} \text{ for } x = 60\º \text{ in the Quadrant I}$

The values where

$\cos(x)=-\frac{1}{2} $   are 120º and 240º.

7 0
3 years ago
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