All categories

3 years ago

If 6t-3=-177 what is the value of the variable

Mathematics

2 answers:

Natalija [7]3 years ago

8 0

6t=-174 (i added three to both sides)
t= -29 (divide by 6 to get your answer)

stealth61 [152]3 years ago

6 0

Add 3 to -3 and -177 then divide -174 by6

You might be interested in

PLEASE HELP PEOPLE!!!! I really need help. urgent. asap.

Zielflug [23.3K]

Answer:

Step-by-step explanation:

8 0

3 years ago

I NEED THIS ASAP about the Pythagorean

Natalija [7]

Answer: 24.49 or 24.5 if rounded.

Step-by-step explanation: We have to use the pythagorean theorem of course...

a² + b² = c²

15² + 20² = c²

225 + 400 = c²

600 = c²

$\sqrt{600}$

c = 24.49 (If we round to the nearest tenth then the answer is 24.5)

8 0

3 years ago

A research group is on a college campus and is conducting a survey regarding the coffee consumption of college students in Ameri

ololo11 [35]

Answer:

it is D because my teacher and I did a sample test and these were one of the questions plus I've seen people with the same answer for this question. take my word for it

Step-by-step explanation:

hope this helps :D (pls brainliest

6 0

2 years ago

Read 2 more answers

What is the product in simplest form x^2+9x+18/x+2 times x^2-3x-10/x^2+2x-24

german

$\dfrac{x^2+9x+18}{x+2}\cdot\dfrac{x^2-3x-10}{x^2+2x-24}=\dfrac{x^2+6x+3x+18}{x+2}\cdot\dfrac{x^2-5x+2x-10}{x^2+6x-4x-24}\\\\=\dfrac{x(x+6)+3(x+6)}{x+2}\cdot\dfrac{x(x-5)+2(x-5)}{x(x+6)-4(x+6)}$

$=\dfrac{(x+6)(x+3)}{x+2}\cdot\dfrac{(x-5)(x+2)}{(x+6)(x-4)}=\dfrac{(x+3)(x-5)}{x-4}\\\\=\dfrac{x^2-5x+3x-15}{x-4}=\dfrac{x^2-2x-15}{x-4}$

5 0

3 years ago

Read 2 more answers

) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

The surface (call it $S$ ) is a triangle with vertices at the points

$x=0,y=0\implies z=2\implies(0,0,2)$

$x=0,y=2\implies z=-2\implies(0,2,-2)$

$x=2,y=0\implies z=-8\implies(2,0,-8)$

Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=(20v,8v,4v)$

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

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv$

$=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}$

8 0

3 years ago