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3 years ago

Help with #6 with work pls

Mathematics

2 answers:

Anna11 [10]3 years ago

7 0

Scale of model is 1 inch represent how many feet in real life
8inch=scale
30 feet=real
8:30=4:15=2:7.5=1:3.75
the one we can se taht matches is D
4 inches=15 feet
D

7. group and add like terms
sum means add btw
5x-2+x+8
5x+x-2+8
6x+6
A

6. D
7. A

kirill115 [55]3 years ago

3 0

D for number 6 and 7 is c

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In right ∆LMN, sin M = 0.759. What is cos N? A. 0.121 B. 0.241 C. 0.349 D. 0.651 E. 0.759

Hatshy [7]

Answer:

E. 0.759

Step-by-step explanation:

You can take this right triangle to have a base length side LM , height of LN and a hypotenuse of MN

The sine of angle ∠LMN is 0.759, find the value of ∠LMN

$sin^-1(0.759)=49.38$

∠LMN=49.38°

Find angle ∠LNM

You know sum of angles in a triangle add up to 180°, given that this is a right-triangle, the base angle is 90° hence

∠LNM=180°-(90°+49.38°)

∠LNM= 180°-139.38°=40.62°

Find cos 40.62°

Cos 40.62°=0.7590

6 0

3 years ago

Help on this math question ‍♀️

noname [10]

Answer:

the 2nd 3rd and 4th answers are the correct ones

Step-by-step explanation:

hope this helps you

4 0

2 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago

15a+2-7<br> What are the terms, like terms, coefficients, and constants in this equation?

Elenna [48]

15 is coefficient, a is a variable, 2 and -7 are like terms, 15a is a term and 2 and 7 are constants.

5 0

2 years ago

Which of the following is an example of the difference of two squares

True [87]

Complete Question: Which of the following is an example of the difference of two squares?

A x² − 9

B x³ − 9

C (x + 9)²

D (x − 9)²

Answer:

A. $x^2 - 9$ .

Step-by-step explanation:

An easy way to spot an expression that is a difference of two squares is to note that the first term and the second term in the expression are both perfect squares. Both terms usually have the negative sign between them.

Thus, difference of two squares takes the following form: $a^2 - b^2 = (a + b)(a - b)$ .

a² and b² are perfect squares. Expanding $(a + b)(a - b)$ will give us $a^2 - b^2$ .

Therefore, an example of the difference of two squares, from the given options, is $x^2 - 9$ .

$x^2 - 9$ can be factorised as $x^2 - 3^2 = (x + 3)(x - 3)$ .

8 0

3 years ago