All categories

3 years ago

HELP PLS FIRST RIGHT ANSWER BRAINLIEST

Mathematics

1 answer:

harkovskaia [24]3 years ago

8 0

okay, so.

it's a rectangle with a nub. We do the area without the nub first. so 8x16 is 128

now, its time for the nub.

6x12-8, that would be 4, so 6x4 is 24+128= 152

So it's B.

Hope i helped Mark brainliest pls

Chad :P

You might be interested in

The sign outside Mrs. Washington’s catering company is in the shape of a trapezoid as shown below what is the permitter of the s

Alona [7]

The perimeter is 2.5+ 2.5+2+3.5= 10.5 ft

The area is equal 6. I used the equation A= a=b/2*h

Hope this helped! Let me know if I can be of more assistance. <span />

3 0

3 years ago

Aditya installed solar panels for her house. When the solar panels produce more electricity than Aditya uses in a day, her net e

Sidana [21]

Answer:

True and false

Step-by-step explanation:

The highest net electricity= true

Net electricity used on july 2nd= false

3 0

3 years ago

How do I apply the change of base formula to y=log (sub 3) x

Ludmilka [50]

$y=\log_3x \Leftrightarrow y=\dfrac{\log_cx}{\log_c3}\text{ where } c\in \mathbb{R}_+\setminus\{1\}$

8 0

3 years ago

Which answers choice shows that the set of irrational numbers is not closed under addition?

babunello [35]

Problably A !!!!!!!!!!!

7 0

3 years ago

Read 2 more answers

Solve this system by elimination : -5x + 3z =8 x- 4y = 7 5y - 4z = -14 find the x , y and z using elimination

Mnenie [13.5K]

Answer

• x = –1

• y = –2

• z = 1

Explanation

Given the system:

$\begin{gathered} -5x+3z=8\text{, equation 1} \\ 5y-4z=-14,\text{ equation 2} \\ x-4y=7,\text{ equation 3} \end{gathered}$

We have to divide equation 1 over 5 and add it to equation 2:

$\begin{gathered} -(5x)+0y+3z=8 \\ x-4y+0z=7 \\ 0x+5y-4z=-14 \end{gathered}$ $\begin{gathered} \frac{-(5x)+0y+3z=8}{5} \\ x-4y+0z=7 \\ 0x+5y-4z=-14 \end{gathered}$ $\begin{gathered} -x+0y+\frac{3}{5}z=\frac{8}{5} \\ x-4y+0z=7 \\ 0x+5y-4z=-14 \end{gathered}$

Now, we have to add 1/5(equation 1) to equation 2:

$\begin{gathered} -x+0y+\frac{3}{5}z=\frac{8}{5} \\ x-4y+0z=7 \\ --------- \\ 0x-4y+\frac{3}{5}z=\frac{43}{5} \end{gathered}$

Next, we multiply the equation 2 obtained previously times 5:

$\begin{gathered} (0x-4y+\frac{3}{5}z=\frac{43}{5})\cdot5 \\ -20y+3z=43 \end{gathered}$

Then, we divide equation 2 over 4:

$\begin{gathered} \frac{-20y+3z=43}{4} \\ -5y+\frac{3}{4}z=\frac{43}{4} \end{gathered}$

We add it to equation 3:

$\begin{gathered} -5y+\frac{3}{4}z=\frac{43}{4} \\ 5y-4z=-14 \\ ------- \\ 0y-\frac{13}{4}z=-\frac{13}{4} \end{gathered}$

Then, we are left with:

$\begin{equation*} -\frac{13}{4}z=-\frac{13}{4} \end{equation*}$

Simplifying:

$-13z=-13$ $z=\frac{-13}{-13}=1$ $z=1$

Now that we have the value of z (z = 1), we can replace it in the modified equation 2 and solve for y:

$-20y+3z=43$ $-20y+3(1)=43$ $-20y+3=43$ $-20y=43-3$ $-20y=40$ $y=\frac{40}{-20}$ $y=-2$

Finally, calculating the value of x with any of the equations (as we already have the other two values):

$x-4(-2)=7$ $x+8=7$ $x=7-8$ $x=-1$

5 0

2 years ago