All categories

2 years ago

WILL GIVE BRAINLIEST ANSWER WHEN POSSIBLE (IF CORRECT)

Mathematics

2 answers:

STALIN [3.7K]2 years ago

5 0

Is there a picture to go with this problem?

uysha [10]2 years ago

3 0

Answer:The answer is C

Step-by-step explanation:

Reflection about the y-axis followed by a translation right by 2 units

You might be interested in

A trapezoid has an area of 36 square feet. If its height is 8, and one base is 4, what is the length of the other base?

Georgia [21]

Answer:

Step-by-step explanation:

A=a+b/2 x h

4 = a

5 = b

8 = h

4 + 5 / 2 = 4.5

4.5 x 8 = 36

So the other base length is 5

5 0

2 years ago

The population of a city decreases by 1.2% per year. If this year's population is 177,000, what will next year's population be,

Vsevolod [243]

Answer:

Initial value

177000

Final value

174,876

Decrease

1.2%

Difference

-2,124

5 0

2 years ago

In ABC AD and BE intersect each other at point q according to a theorem on medians q divides AD in the ratio 2:1 what are the co

victus00 [196]

C. (3.33, -3) would be the answer

8 0

2 years ago

1 Each tire on Bruno's delivery truck weighs 48.25

Vitek1552 [10]

Answer: A. 289.5

Method: 48.25 x 6 = 289.5

3 0

3 years ago

Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?

Ede4ka [16]

Answer:

The solution is $x=2,\ y=0,\ z=-1.$

Step-by-step explanation:

You are given the system of three equations:

$\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.$

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

$\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.$

So,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.$

Multiply the third equation by 23 and subtract it from the second equation:

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.$

Hence,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\14z=-14 \end{array}\right.\Rightarrow z=-1$

Substitute it into the second equation:

$23y-9\cdot (-1)=9\Rightarrow 23y+9=9\\ \\23y=0\\ \\y=0$

Substitute them into the first equation:

$3x+2\cdot 0+3\cdot (-1)=3\Rightarrow 3x-3=3\\ \\3x=6\\ \\x=2$

The solution is $x=2,\ y=0,\ z=-1.$

3 0

3 years ago