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Nikitich [7]
3 years ago
15

Simplify 4^2 ⋅ 4^8. A)4^16 B)4^10 C)1^616 D)1^610

Mathematics
1 answer:
Wewaii [24]3 years ago
6 0
4^2 × 4^8 =
= 4^2+8
= 4^10
So, answer is B.
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If an airplane in a 1 in. : 4 ft. scale drawing is 5 in. long, how long is the actual airplane? Use proportions to solve the pro
Lunna [17]

Answer:

20ft

Step-by-step explanation:

20ft is the answer because you would do 4 x 5 = 20, since the scale factor is 1in : 4ft

Hope this helped :)

4 0
3 years ago
How many soultions? -4x-7+10x=-7+6x
hammer [34]

Answer:

C. Infinitely many solutions.

Step-by-step explanation:

-4x - 7 + 10x = -7 + 6x

Combine like terms on the left side. Rearrange the right side.

6x - 7 = 6x - 7

Add 7 to both sides.

6x = 6x

Subtract 6x from both sides.

0 = 0

0 = 0 is a true statement.

Both sides are equal, so all real values of x make the equation true.

Answer: C. Infinitely many solutions.

4 0
3 years ago
The population of a town can be modeled by P= 22,000 + 125t, where P represents the population and the t represents the number o
nlexa [21]
Given that the population can be modeled by P=22000+125t, to get the number of years after which the population will be 26000, we proceed as follows:
P=26000
substituting this in the model we get:
26000=22000+125t
solving for t we get:
t=4000/125
t=32
therefore t=32 years
This means it will take 32 years for the population to be 32 years. Thus the year in the year 2032
7 0
3 years ago
What is the approximate area of the triangle below?
Alecsey [184]
Can you include a picture of the problem theres no information
7 0
3 years ago
Read 2 more answers
Verify identity: <br><br> (sec(x)-csc(x))/(sec(x)+csc(x))=(tan(x)-1)/(tan(x)+1)
Nikitich [7]
So hmmm let's do the left-hand-side first

\bf \cfrac{sec(x)-csc(x)}{sec(x)+csc(x)}\implies \cfrac{\frac{1}{cos(x)}-\frac{1}{sin(x)}}{\frac{1}{cos(x)}+\frac{1}{sin(x)}}\implies &#10;\cfrac{\frac{sin(x)-cos(x)}{cos(x)sin(x)}}{\frac{sin(x)+cos(x)}{cos(x)sin(x)}}&#10;\\\\\\&#10;\cfrac{sin(x)-cos(x)}{cos(x)sin(x)}\cdot \cfrac{cos(x)sin(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}

now, let's do the right-hand-side then  

\bf \cfrac{tan(x)-1}{tan(x)+1}\implies \cfrac{\frac{sin(x)}{cos(x)}-1}{\frac{sin(x)}{cos(x)}+1}\implies \cfrac{\frac{sin(x)-cos(x)}{cos(x)}}{\frac{sin(x)+cos(x)}{cos(x)}}&#10;\\\\\\&#10;\cfrac{sin(x)-cos(x)}{cos(x)}\cdot \cfrac{cos(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}

7 0
3 years ago
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