What’s the answer to this
What’s the answer to this - 1

Answers

Answer 1
Answer: answer : x-intercept = (11,0)
y-intercept = (0,6)

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What is the solution to x+5>-7

Answers

According to the given data we have the following inequality:

x+5>-7

To find the solution of the inequality above we would make the following:

\begin{gathered} \mathrm{Subtract\: }5\mathrm{\: from\: both\: sides} \n x+5-5>-7-5 \end{gathered}\begin{gathered} \text{Finally, simplify},\text{ and so:} \n x>-12 \end{gathered}

Therefore, the solution to x+5>-7​ would be x>-12

Students are asked to memorize a list of 100 words. The students are given periodic quizzes to see how many words they have memorized. The function L gives the number of words memorized at time t. The rate of change of the number of words memorized is proportional to the number of words left to be memorized. 1. Which of the following differential equations could be used to model this situation, where k is a positive constant?
A. dL/dt = kL
B. dL/dt = 100 - kL
C. dL/dt = k(100 - L)
D. dL/dt = kL - 100

Answers

Answer:

C. dL/dt = k(100 - L)

Step-by-step explanation:

We have a list containing 100 words.

L = number of words memorized at time t.

At any time t, the number of words left to be memorized is 100-L.

Therefore:

(dL)/(dt)\propto 100-L\n $Introducing k, a constant  of proportion$\n(dL)/(dt)= k(100-L)

The correct option is C.

The weights of adobe bricks used for construction are normally distributed with a mean of 3 pounds and a standard deviation of 0.25 pound. Assume that the weights of the bricks are independent and that a random sample of 28 bricks is selected. Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that all the bricks in the sample exceed 2.75 pounds? (b) What is the probability that the heaviest brick in the sample exceeds 3.75 pounds?

Answers

Answer:  a) 0.8413, b) 0.9987.

Step-by-step explanation:

Since we have given that

Mean = 3 pounds

Standard deviation = 0.25 pounds

n = 28 bricks

So, (a) What is the probability that all the bricks in the sample exceed 2.75 pounds?

P(X>2.75)\n\n=P(z>(2.75-3)/(0.25)\n\n=P(z>(-0.25)/(0.25))\n\n=P(z>-1)\n\n=0.8413

b) What is the probability that the heaviest brick in the sample exceeds 3.75 pounds?

P(X>3.75)\n\n=P(z>(3.75-3)/(0.25))\n\n=P(z>(0.75)/(0.25))\n\n=P(z>3)\n\n=0.9987

Hence, a) 0.8413, b) 0.9987.

Use the binomial theorem to expand the expression :(3x + y)^5 and simplify.
(b) find the middle term in the expansion of
(1/x+√x)^4 and simplify your unswer.
(c) determine the coefficient of x^11 in the expansion of (x^2 +1/x)^10, simplify your answer.

Answers

Answer:

a) (3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5.

b) The middle term in the expansion is (6)/(x).

c) The coefficient of x^(11) is 120.

Step-by-step explanation:

Remember that the binomial theorem say that (x+y)^n=\sum_(k=0)^(n) \binom{n}{k}x^(n-k)y^(k)

a) (3x+y)^5=\sum_(k=0)^5\binom{5}{k}3^(n-k)x^(n-k)y^k

Expanding we have that

\binom{5}{0}3^5x^5+\binom{5}{1}3^4x^4y+\binom{5}{2}3^3x^3y^2+\binom{5}{3}3^2x^2y^3+\binom{5}{4}3xy^4+\binom{5}{5}y^5

symplifying,

(3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5.

b) The middle term in the expansion of ((1)/(x) +√(x))^4=\sum_(k=0)^(4)\binom{4}{k}(1)/(x^(4-k))x^{(k)/(2)} correspond to k=2. Then \binom{4}{2}(1)/(x^2)x^{(2)/(2)}=(6)/(x).

c) (x^2+(1)/(x))^(10)=\sum_(k=0)^(10)\binom{10}{k}x^(2(10-k))(1)/(x^k)=\sum_(k=0)^(10)\binom{10}{k}x^(20-2k)(1)/(x^k)=\sum_(k=0)^(10)\binom{10}{k}x^(20-3k)

Since we need that 11=20-3k, then k=3.

Then the coefficient of x^(11) is \binom{10}{3}=120

A pool in the shape of a rectangular prism is 6 meters long and 3 meters wide. the water in the pool is 1 meter deep.a. the density of water is about 1 gram per cubic centimeter. find the number of kilograms of water in the pool. question 2
b. you add 6000 kilograms of water to the pool. what is the depth of the water in the pool? write your answer as a fraction. the water is about meters deep.

Answers

Answer:

1). Mass of water present= 18000 kg

2).4/3 meters deep

Step-by-step explanation:

Area of the rectangle= 6*3= 18m²

Volume of water in the pool

= Deepness of water*area of rectangle

= 1*18

= 18 m³

density of water is about 1 gram per cubic centimeter

In kg per m³= 1000 kg/me

Mass of water present= density*volume

Mass of water present= 1000*18

Mass of water present= 18000 kg

2)6000 kilograms of water is added to 18000 of

Total mass present= 6000+18000

Total mass present=24000 kg

If density= 1000kg/m³

Volume present= mass/density

Volume present= 24000/1000

Volume present= 24 m³

Area of the rectangle= 18 m²

deepness of the pool= volume/area

deepness of the pool= 24/18

deepness of the pool= 4/3 meters deep

Which two numbers both round to 1,500 when rounded to the nearest hundred?

Answers

The two numbers where  both round to 1,500 when rounded to the nearest hundred is option c. 1457 and 1547.

What is Rounded number?

The rounded number should be that where the number should be rounded to the nearest tenth, hundred or thousand

Since the number is 1500

So for hundred the number should be likely 1457 and 1547.

hence, the option c is correct.

learn more about number here: brainly.com/question/21474983

Answer:

C

Step-by-step explanation:

Just take it lol