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2 years ago

A picture of two horses pulling a carriage to the left.

Mathematics

1 answer:

IrinaVladis [17]2 years ago

7 0

Answer:

1,327

Step-by-step explanation:

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Multiply 5,5,5,5 4 times which gives me 625

k0ka [10]

Answer:

Confusion

Step-by-step explanation:

what the heck are you asking?

4 0

2 years ago

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Find f(g(x)) and g(f(x)).<br> f(x)=x²-2 and g(x)=√x+1

Archy [21]

Answer:

1. x - 1

2. $\sqrt{x^2-1}$

Step-by-step explanation:

$f(g(x)) = \sqrt{x+1}^2 - 2\\ = x + 1 - 2\\= x - 1$

$g(f(x)) =\sqrt{x^2-2+1}\\ = \sqrt{x^2-1}$

6 0

1 year ago

On the first part of her trip Natalie rode her bike 16 miles and on the second part of the trip she rode her bike 42 miles. Her

-BARSIC- [3]

Answer:

14 mph ( average speed during the second part of the trip )

Step-by-step explanation:

Let´s call "x" the average speed during the first part then

t = 5 hours

t = t₁ + t₂ t₁ and t₂ times during part 1 and 2 respectively

l = t*v ( distance is speed by time ) t = l/v

First part

t₁ = 16/x and t₂ = 42 / ( x + 6)

Then

t = 5 = 16/x + 42 /(x + 6)

5 = [ 16 * ( x + 6 ) + 42 * x ] / x* ( x + 6 )

5 *x * ( x + 6 ) = 16*x + 96 + 42 x

5*x² + 30*x - 58*x - 96 = 0

5*x² - 28*x - 96 = 0

We obtained a second degree equation, we will solve for x and dismiss any negative root since negative time has not sense

x₁,₂ = [28 ± √ (28)² + 1920 ] / 10

x₁,₂ = ( 28 ± √2704 )/ 10

x₁ = 28 - 52 /10 we dismiss that root

x₂ = 80/10

x₂ = 8 mph average speed during the first part, and the average speed in the second part was 6 more miles than in the firsst part. then the average spedd dring the scond part was 8 + 6 = 14 mph

6 0

3 years ago

Simplify.<br> 10u -4u<br> Please help

vovikov84 [41]

Answer:

Step-by-step explanation:

The two terms 10u and 4u are like terms because they both carry the same variable (u) with the same degree (1). So, we can just subtract them as if they were normal numbers 10 - 4:

10u - 4u = 6u

Thus, the answer is 6u.

Hope this helps!

8 0

3 years ago

Read 2 more answers

Rectangle ABCD was dilated to create rectangle A'B'C'D.

77julia77 [94]

The first thing we must do for this case is to calculate the scale factor.

For this, we make the relationship between two parallel sides.

We have then:

$k =\frac{B'C'}{BC}$