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4 years ago

Please help me find surface area for each. Round to the nearest tenth if needed.

Mathematics

1 answer:

Degger [83]4 years ago

3 0

Answer:

168 in², 94.25 yd²

Step-by-step explanation:

I couldn't find a formula for the SA of a triangular prism so I just found the area of each surface.

The formula for SA of a cylinder is $2\pi rh+2\pi r^{2}$

All work for triangular prism is in the image

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HELP MATH ASAP IF YOU PUT IN A RANDOM ANSWER YOU WILL BE REPORTED TYSM!

VikaD [51]

Answer:

Hello! Mizuki here to help!

The correct answer for problem 9 is B.

The correct answer for problem 10 is also B.

Step-by-step explanation:

Sorry if the work I did is confusing, if you don't get what I did, just comment below and I'll try my best to answer your question!

problem 9 work:

3 x 4 + 5 x 8 + 4 x 8 + 3 x 8 =

12 + 40 + 32 + 24 =

108

problem 10 work:

17 x 24 + 8 x 15 + 15 x 24 + 8 x 24 =

408 + 120 + 360 + 192 =

1080

7 0

3 years ago

Read 2 more answers

How many phone numbers are possible in the (770) area code if: For the form ABC-XXXX, A is restricted to numbers 2-9. B, C, and

weqwewe [10]

Answer:

c.7,999,999

Step-by-step explanation:

The phone number is of the form ABC - XXXX

A can be any number from 2 - 9. This means number of possible values for A are 8.

The rest of the places B,C and X can be any digit from 0 - 9. This means there are 10 possible values for each of these.

Since, value to A can be assigned in 8 ways, and to the rest of the 6 positions in 10 ways, according to the fundamental rule of counting, the total number of possible phone numbers that can be formed will be equal to the product of all the individual ways:

Total possible phone numbers = 8 x 10 x 10 x 10 x 10 x 10 x 10

Since, 1 of the given number: 867-5309 is not used, the total possible phone numbers will be:

Total possible phone numbers = $8 \times 10^{6} - 1 = 7999999$

Hence, option C: 7,999,999 give the correct answer.

6 0

3 years ago

HELP MEeeeeeeeee g: R² → R a differentiable function at (0, 0), with g (x, y) = 0 only at the point (x, y) = (0, 0). Consider<im

GrogVix [38]

(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.

• Recall that for $f:\mathbb R^2\to\mathbb R$ , we have the partial derivative with respect to $x$ defined as

$\displaystyle \frac{\partial f}{\partial x} = \lim_{h\to0}\frac{f(x+h,y) - f(x,y)}h$

The derivative at (0, 0) is then

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(0+h,0) - f(0,0)}h$

• By definition of $f$ , $f(0,0)=0$ , so

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)}h = \lim_{h\to0}\frac{\tan^2(g(h,0))}{h\cdot g(h,0)}$

• Expanding the tangent in terms of sine and cosine gives

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{h\cdot g(h,0) \cdot \cos^2(g(h,0))}$

• Introduce a factor of $g(h,0)$ in the numerator, then distribute the limit over the resulting product as

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{g(h,0)^2} \cdot \lim_{h\to0}\frac1{\cos^2(g(h,0))} \cdot \lim_{h\to0}\frac{g(h,0)}h$

• The first limit is 1; recall that for $a\neq0$ , we have

$\displaystyle\lim_{x\to0}\frac{\sin(ax)}{ax}=1$

The second limit is also 1, which should be obvious.

• In the remaining limit, we end up with

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)}h = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h$

and this is exactly the partial derivative of $g$ with respect to $x$ .

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h = \frac{\partial g}{\partial x}(0,0)$

For the same reasons shown above,

$\displaystyle \frac{\partial f}{\partial y}(0,0) = \frac{\partial g}{\partial y}(0,0)$

(b) To show that $f$ is differentiable at (0, 0), we first need to show that $f$ is continuous.

• By definition of continuity, we need to show that

$\left|f(x,y)-f(0,0)\right|$

is very small, and that as we move the point $(x,y)$ closer to the origin, $f(x,y)$ converges to $f(0,0)$ .

We have

$\left|f(x,y)-f(0,0)\right| = \left|\dfrac{\tan^2(g(x,y))}{g(x,y)}\right| \\\\ = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)^2}\cdot\dfrac{g(x,y)}{\cos^2(g(x,y))}\right| \\\\ = \left|\dfrac{\sin(g(x,y))}{g(x,y)}\right|^2 \cdot \dfrac{|g(x,y)|}{\cos^2(x,y)}$

The first expression in the product is bounded above by 1, since $|\sin(x)|\le|x|$ for all $x$ . Then as $(x,y)$ approaches the origin,

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{|g(x,y)|}{\cos^2(x,y)} = 0$

So, $f$ is continuous at the origin.

• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which $f(x,y)$ changes as we move the point $(x,y)$ closer to the origin, given by

$\left|\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\right|,$

approaches 0.

Just like before,

$\left|\dfrac{\tan^2(g(x,y))}{g(x,y)\sqrt{x^2+y^2}}\right| = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)}\right|^2 \cdot \left|\dfrac{g(x,y)}{\cos^2(g(x,y))\sqrt{x^2+y^2}}\right| \\\\ \le \dfrac{|g(x,y)|}{\cos^2(g(x,y))\sqrt{x^2+y^2}}$