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2 years ago

PLEASE HELP ME SOLVE THESE QUESTIONSIT WOULD BE MUCH APPRECIATED

Mathematics

1 answer:

Ugo [173]2 years ago

6 0

Answer:

1. b.

2. a.

3. b.

4. d.

5. a.

6. b.

7. c.

8. b.

9. b.

10. c.

11. a.

12. b.

13. a.

14. c.

Step-by-step explanation:

7 + 3 = 10

+ 679 - 2056 = -1377

- 398 + 1 = -397

-23 - 237 = 260

2 - 6 = -4

-2 × 3 × -4 = 24

absolute value of -36 = 36

6 + 4 = 10

- 5928 - 965 = -6893

215 × 212 + 212 × 314 = 112148

(15 + 5) ÷ (15 + 5) = 20 ÷ 20 = 1

-457 + 1 = -456

- 218 -1 = -219

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15x + 3 + 2x + 9r - (8 - r)

bezimeni [28]

Answer:

17x + 10r -5

Step-by-step explanation:

15x +3 + 2x + 9r- (8-r)

17x +3 +9r -8 +r

17x+ 10r+ -5

6 0

3 years ago

Clara is growing her hair out and wants it to be no longer than 15 cm but more than 11 cm. Her hair is 7 cm right now and grows

sertanlavr [38]

For this case we have the following compound inequality:
11 <7 + 0.25t ≤ 15
We must solve this inequality for time.
We have then:

For 11 <7 + 0.25t:
11 <7 + 0.25t
11 - 7 <0.25t
4 <0.25t
4 / 0.25 <t
16 <t

For 7 + 0.25t ≤ 15:
7 + 0.25t ≤ 15
0.25t ≤ 15 - 7
0.25t ≤ 8
t ≤ 8 / 0.25
t ≤ 32

Therefore, the answer is:
16 <t ≤ 32

Answer:
16 <t ≤ 32
Clara has the desired length in more than 16 weeks, but up to 32 weeks.

6 0

3 years ago

Clara writes the following proof for the theorem: If the diagonals of a quadrilateral bisect each other, the quadrilateral is a

zheka24 [161]

Answer:

CPCTC

Step-by-step explanation:

Statements 3 and 4 show the top and bottom triangles are congruent, and the left and right triangles are congruent. Statement 5 is making use of these facts to claim that the alternate interior angles are congruent. This claim is valid because ...

Corresponding Parts of Congruent Triangles are Congruent (CPCTC)

7 0

2 years ago

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in the solution. Water containing1 lb

devlian [24]

Answer:

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is $\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$ .

(b) The concentration (in lbs per gallon) when it is at the point of overflowing is $\frac{121}{125}\:\frac{lb}{gal}$ .

(c) The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$ .

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of Q(t) = Rate at which Q(t) enters the tank – Rate at which Q(t) exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x (concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x (concentration of substance in liquid exiting)

Let $C$ be the concentration of salt water solution in the tank (in $\frac{lb}{gal}$ ) and $t$ the time (in minutes).

Since the solution being pumped in has concentration $1 \:\frac{lb}{gal}$ and it is being pumped in at a rate of $3 \:\frac{gal}{min}$ , this tells us that the rate of the salt entering the tank is

$1 \:\frac{lb}{gal} \cdot 3 \:\frac{gal}{min}=3\:\frac{lb}{min}$

But this describes the amount of salt entering the system. We need the concentration. To get this, we need to divide the amount of salt entering the tank by the volume of water already in the tank.

$V(t)$ is the volume of brine in the tank at time t. To find it we know that at t = 0 there were 200 gallons, 3 gallons are added and 2 are drained, and the net increase is 1 gallons per second. So,

$V(t)=200+t$

Therefore,

The rate at which $C(t)$ enters the tank is

$\frac{3}{200+t}$

The rate of the amount of salt leaving the tank is

$C\:\frac{lb}{gal} \cdot 2 \:\frac{gal}{min}+C\:\frac{lb}{gal} \cdot 1\:\frac{gal}{min}=3C\:\frac{lb}{min}$

and the rate at which $C(t)$ exits the tank is

$\frac{3C}{200+t}$

Plugging this information in the main equation, our differential equation model is:

$\frac{dC}{dt} =\frac{3}{200+t}-\frac{3C}{200+t}$

Since we are told that the tank starts out with 200 gal of solution, containing 100 lb of salt, the initial concentration is

$\frac{100 \:lb}{200 \:gal} =0.5\frac{\:lb}{\:gal}$

Next, we solve the initial value problem

$\frac{dC}{dt} =\frac{3-3C}{200+t}, \quad C(0)=\frac{1}{2}$

$\frac{dC}{dt} =\frac{3-3C}{200+t}\\\\\frac{dC}{3-3C} =\frac{dt}{200+t} \\\\\int \frac{dC}{3-3C} =\int\frac{dt}{200+t} \\\\-\frac{1}{3}\ln \left|3-3C\right|=\ln \left|200+t\right|+D\\\\$

We solve for C(t)

$C(t)=1+D(200+t)^{-3}$

D is the constant of integration, to find it we use the initial condition $C(0)=\frac{1}{2}$

$C(0)=1+D(200+0)^{-3}\\\frac{1}{2} =1+D(200+0)^{-3}\\D=-4000000$

So the concentration of the solution in the tank at any time t (before the tank overflows) is

$C(t)=1-4000000(200+t)^{-3}$

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is just the concentration of the solution times its volume

$(1-4000000(200+t)^{-3})(200+t)\\\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$

(b) Since the tank can hold 500 gallons, it will begin to overflow when the volume is exactly 500 gal. We noticed before that the volume of the solution at time t is $V(t)=200+t$ . Solving the equation

$200+t=500\\t=300$

tells us that the tank will begin to overflow at 300 minutes. Thus the concentration at that time is

$C(300)=1-4000000(200+300)^{-3}\\\\C(300)= \frac{121}{125}\:\frac{lb}{gal}$

(c) If the tank had infinite capacity the concentration would then converge to,

$\lim_{t \to \infty} C(t)= \lim_{t \to \infty} 1-4000000\left(200+t\right)^{-3}\\\\\lim _{t\to \infty \:}\left(1\right)-\lim _{t\to \infty \:}\left(4000000\left(200+t\right)^{-3}\right)\\\\1-0\\\\1$

The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$

4 0

3 years ago

What Should i write down basically i just need the answers and let me copy and paste

Anastaziya [24]

Answer:

✔️Ratio of weight to mass in simplest form: 49/5 or 49 : 5

✔️x = 110

Step-by-step explanation:

Since the ratio of weight to mass is constant, therefore we can use any pair of values from the table to find the ratio that represents the table. Let's pick the first row:

(20, 196)

Ratio of weight to mass = 196 : 20

= 196/20

= 49/5

Or 49 : 5

✔️Let's use this ratio/proportion to find x in the table

Ratio of weight to mass = 49:5

Thus:

1078/x = 49/5

Cross multiply

x*49 = 1078*5

49x = 5,390

x = 5,390/49

x = 110

5 0

2 years ago

PLEASE HELP ME SOLVE THESE QUESTIONSIT WOULD BE MUCH APPRECIATED ​

PLEASE HELP ME SOLVE THESE QUESTIONSIT WOULD BE MUCH APPRECIATED