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2 years ago

Please help me !! is/is notcan/cannotis/is notcan/cannot

Mathematics

1 answer:

zlopas [31]2 years ago

7 0

Answer:

isn't

can't

Step-by-step explanation:

hjjhfyuhvdgujbcff

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Graph the equations to solve the system y=x+4 y= -x - 6

Mama L [17]

Answer:

(-5, -1)

General Formulas and Concepts:

<u>Algebra I</u>

Functions

Graphing

Reading a Cartesian Plane
Coordinates (x, y)

Solving systems of equations by graphing

Step-by-step explanation:

Where the 2 equations intersect is the solution to the systems of equations.

5 0

3 years ago

un cine tiene 2 salas la sala A tiene 1920 butacas distribuidas en 24 filas con la misma cantidad en cada una ¿cuantas butacas t

Svetlanka [38]

Answer:

80 and B is 96

Step-by-step explanation:

1920 ÷ 24 = 80

1920 ÷ 20 = 96

puedes comprobarlo en la calculadora

lo siento si es incorrecto

eso es lo que entendí

3 0

3 years ago

jan, mya, and sara ran a total of 64 miles last week. jan and mya ran the same number of miles. sara ran 8 less miles than jan a

kherson [118]

$x-\ Jan's\ destination\\\\y-\ Mya's \ destination \\\\ z-\ Sara's\ destination\\\\\
x+y+z=64\\
x=y\\
z-8=2x\\\\
x+x+z=64\\
z=2x+8\\\\
2x+2x+8=64\\\\
4x+8=64\\\\
4x=64-8\\\\
4x=56\ \ |:4\\\\x=14\\\\z=2x+8=2*14+8=28+8=36\\\\ Sara\ ran\ 36\ miles.$

5 0

3 years ago

HELP PLZZZZZZZZZZZZ!!!!!!!!!!!!!!!! <br> ILL GIVE YOU 15 POINTS

seropon [69]

Answer:

Step-by-step explanation:

a rational number is a number that can be expressed as a fraction p/q of two integers, q cannot be 0

so for A

$\sqrt{4/5}$ Cannot be expressed as 2 integers as a quotient

so A is wrong

For B .125 is 1/8 so yes -2 is a integer so yes 2/5 are 2 integers so yes

and $\sqrt{16/9}$ Is 4/3 so yes

7 0

3 years ago

Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]

Anit [1.1K]

Answer:

The answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

Step-by-step explanation:

$\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}$

Solve the L.H.S part:

$\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]$

After calculating the L.H.S part compare the value with R.H.S:

$\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\$

$\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\$

In equation (i) multiply by 3 and subtract by equation (iii):

$\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to c= - \frac{1}{2}$

put the value of c in equation (i):

$\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\$

In equation (ii) multiply by 3 then subtract by equation (iv):

$\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\$

put the value of d in equation (iv):

$\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2$

The final answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

4 0

3 years ago

Please help me !! is/is notcan/cannotis/is notcan/cannot​

Please help me !! is/is notcan/cannotis/is notcan/cannot