Ask question

Ask question

All categories

3 years ago

7

Simplify the expressing by combining like terms 1 + 3x - 8x + 7

1 answer:

tamaranim1 [39]3 years ago

7 0

$\huge\text{Hey there!}$

$\large\textsf{1 + 3x - 8x + 7}$

$\large\text{COMBINE the LIKE TERMS }$

$\large\textsf{ (3x - 8x) + (1 + 7)}$

$\large\textsf{3x - 8x = \bf -5x}$

$\large\textsf{1 + 7 = \bf 8}$

$\large\textsf{= -5x + 8}$

$\boxed{\boxed{\large\text{Answer: \huge \bf -5x + 8}}}\huge\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

You might be interested in

-2×-7y=30 <br>7×+4y=18 <br>×= <br>y=

Sati [7]

Answer:

Answer is -7 Hope this helps :D

Step-by-step explanation:

6 0

3 years ago

Does anyone know this

irga5000 [103]

Answer:

The scale factor is greater than 1

Step-by-step explanation:

7 0

3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

Jayden painted 1/6 of a wall in 2/3 hour. At this rate , how many hours will it take Jayden to paint the wall.

Anni [7]

Answer:

Step-by-step explanation:

4 0

3 years ago

Natalie consumes only apples and tomatoes. Her utility function is U(x, y) = x 2y 8 , where x is the number of apples consumed a

Kamila [148]

Answer:

16 apples.

Step-by-step explanation:

$U(x, y) = x^2y^8$

$MU_x =\frac{\partial U}{\partial x} = 2xy^8\\MU_y = \frac{\partial U}{\partial x} = 8x^2y^7\\$

Price of apples, Px=$4

Price of tomatoes, Py=$3

Ratio of their Marginal Utilities

$\frac{MU_x}{MU_y} = \frac{Px}{Py}$

$\frac{y}{4x} = \frac{4}{3}$

$y=\frac{16x}{3}$

Since Natalie’s income is $320

320 = xPx+yPy

$320=4x + 3*\frac{16x}{3} \\320= 4x+16x = 20x\\x=\frac{320}{20} = 16\\Also\\y=\frac{16x}{3}=\frac{16X16}{3}=\$85.33$

Since x=16, Natalie will consume 16 apples.

6 0

3 years ago