All categories

2 years ago

What is an equivalent expression for this:

Mathematics

1 answer:

Veseljchak [2.6K]2 years ago

8 0

25a ^4 b^4 y^10 all of them are multiplied to each other I just separated it so it didn’t look confusing

You might be interested in

Gavin drives at a speed of 70 miles per hour.<br> How long will it take to drive 245 miles?

katen-ka-za [31]

Answer:

he will spend all his money on gas and not make it down the block

3 0

2 years ago

Read 2 more answers

PRACTICE ANOTHER A piece of wire 18 m long is cut into two pieces. One piece is bent into a square and the other is bent into an

sergejj [24]

Answer:

Step-by-step explanation:

We have the equations

4x + 3y = 18 where x = the side of the square and y = the side of the triangle

For the areas:

A = x^2 + √3y/2* y/2

A = x^2 + √3y^2/4

From the first equation x = (18 - 3y)/4

So substituting in the area equation:

A = [ (18 - 3y)/4]^2 + √3y^2/4

A = (18 - 3y)^2 / 16 + √3y^2/4

Now for maximum / minimum area the derivative = 0 so we have

A' = 1/16 * 2(18 - 3y) * -3 + 1/4 * 2√3 y = 0

-3/8 (18 - 3y) + √3 y /2 = 0

-27/4 + 9y/8 + √3y /2 = 0

-54 + 9y + 4√3y = 0

y = 54 / 15.93

= 3.39 metres

So x = (18-3(3.39) / 4 = 1.96.

This is a minimum value for x.

So the total length of wire the square for minimum total area is 4 * 1.96

= 7.84 m

There is no maximum area as the equation for the total area is a quadratic with a positive leading coefficient.

3 0

2 years ago

Which equation has the correct sign on the product

Nata [24]

C. Because c is makes more sense

7 0

2 years ago

Read 2 more answers

Can someone help me round these two numbers? I completely forgot how ‍

tigry1 [53]

Answer:

Step-by-step explanation:

If you are rounding to nearest tenth

2.6 1.9

Nearest hundredth

2.6 1.87

If the number to the right of the one you want to round is 5 or above, you go up a number. If it's below, you go down 1.

Hope this helps.

5 0

2 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

2 years ago