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3 years ago

Can someone help me with this, I will mark you as brainliest!

Mathematics

1 answer:

Aleksandr [31]3 years ago

8 0

The second one is the answer because you have to start with 180 first meaning you rotate 180 and reflect which is rx .hope this helps

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Simplify the expression.

deff fn [24]

(6^-2)^2

When you have a number raised to a power inside parenthesis and then it is raised to another power, to simplify, multiply the two powers together:

-2 x 2 = -4

Simplified is 6^-4

5 0

3 years ago

Read 2 more answers

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$

8 0

3 years ago

What is the least common denominator of the fractions 2/3, 1/4, 1/8 and ?

KiRa [710]

Answer:

Step-by-step explanation:

That's the first number that all your denominators have in common.

3 0

3 years ago

Please help me, and I will mark you as brainliest

Lostsunrise [7]

Answer:

what is the equation that u actually looking for?

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Offering 40 points and brainliest for answer

Nata [24]

In fact, this problem belongs to the chemistry section. Recall that many other sciences require mathematical calculations. The problem will belong to Mathematics only if no knowledge of other sciences are required to solve the problem.

Solubility for the given substances is measured in grams per 100 g of water at a particular temperature (20 deg.C).
This means that the mass (assumed to be the solute) will not change the solubility, just the minimum quantity of solvent (water) will.
Thus the solubility of sodium chloride will remain L=36 g/100g H2O for any quantity of solute. Similarly, the solubility of lead nitrate will remain as K=54 g/100 g H2O.

The reason that they remain constant is because the quantity of solvent (water) is fixed at 100 g. Varying amount of solute will affect the quantity of solvent required, but not the solubility.
I'll leave it to you to calculate the difference between K & L.

8 0

3 years ago