All categories

3 years ago

The sum area please.

Mathematics

1 answer:

Ann [662]3 years ago

3 0

Answer:

4xy+24y^2−7x−38y−7

Step-by-step explanation:

(-7+4y) (6y+x+1)

-7*6y= -42y

-7*x= -7x

-7*1= -7

4y*6y= 24y²

4y*x= 4yx

4x*1= 4y

Add them up

You might be interested in

john borrowed $5 dollars per month from his mom to pay for his art supplies at the end of the course he owed his mom $65 dollars

Kobotan [32]

$65 total ÷ $5 a month = 13 months total

John borrowed money from his mom for 13 months.

5 0

3 years ago

Read 2 more answers

The two expressions below have the same value when rounded to the nearest hundredth. log5 b log9 48 What is the approximate valu

Norma-Jean [14]

Answer:

The answer is the second option: 1.23

Step-by-step explanation:

To solve this problem we use the logarithmic base change property.

Let c, d and h be positive real numbers then:

$log_c(d) = \frac{log_h(d)}{log_h(c)}$

We also know that:

$log_{10}(c)$ is also written as $log(c)$ .

Then, we can write:

$log_5(b) = \frac{log(b)}{log(5)}$

Then:

$log_5(b) = log_9(48)$

So:

$log_{9}(48) = \frac{log(b)}{log(5)}$

$log(b) = log(5)[log_9(48)]\\\\log(b) = 0.69897[1.7619]\\\\log(b) = 1.23$

5 0

3 years ago

Read 2 more answers

Milo cut his cake into 2 equal pieces. He ate 2pieces of cake for his birthday.

Sauron [17]

Answer:

$1$ $or \frac{2}{2}$

Step-by-step explanation:

<h2><u>We know that Milo cut his cake into 2 different pieces, so:</u></h2>

$\frac{x}{2}$

<h2><u>He then eats two pieces of the cake. so:</u></h2>

$2$ $pieces$ $of$ $cake$ $=$ $x$

$\frac{2}{2} = 1$

8 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

4 years ago

If the third term in a geometric sequence is 20, and the fifth term in the

Gwar [14]

Answer: A) 1/2

Step-by-step explanation:

In a geometric sequence, the consecutive terms differ by a common ratio. The formula for determining the nth term of a geometric progression is expressed as

Tn = ar^(n - 1)

Where

a represents the first term of the sequence.

r represents the common ratio.

n represents the number of terms.

If the third term is 20, it means that

T3 = 20 = ar^(3 - 1)

20 = ar²- - - - - - - - - - 1

If the third term is 20, it means that

T5 = 5 = ar^(5 - 1)

5 = ar⁴- - - - - - - - - - 2

Dividing equation 2 by equation 1, it becomes

5/20 = r⁴/r²

1/4 = r^(4 - 2)

(1/2)² = r²

r = 1/2

6 0

4 years ago