Use the function Y=2x^2+3x+7 to answer the following questions

Answer:

M = 4/3 - c/3

Step-by-step explanation:

Answer:

4/3

Step-by-step explanation:

The slope formula is the difference in y over the difference in x, so 4-0/3-0.

Answer:

M=6 OR slope=6

Step-by-step explanation:

Use slope formula

Answer:

Step-by-step explanation:

Slope is defined as the change of y over the change of x or rise over run. This is calculated by subtracting the second set of vales from the first set of values.

For y:

We have 10 and 4 for the y coordinates (y is the second number in a coordinate set). We will subtract 4 from 10, to get a result of 6.

For x:

We have 3 and 0 for the x coordinates (x is the first number in a coordinate set). We will subtract 0 from 3, to get a result of 3.

Slope is 6/3 or 2/1

Answer:

a. What does the Greene family pay as an annual premium? $450.60

b. How much would it cost him to purchase insurance in the amount of 100/300/25? $480.70

c. What is the difference in cost? $30.10

(っ◔◡◔)っ ♥here you go♥

Answer:

he should buy 3 gallons

Step-by-step explanation:

There are 16 cups in a gallon, and if each person will eat 2 cups, and there are 24 ppl, multiply 24 by 2 to get the number of cups.

24 x 2 = 48.

Then, divide that by 16 (the number of cups in a gallon) to t how many gallons.

48 ÷ 16 = 3

so, he needs 3 gallons

hope this helped!  

Answer:

B. 4 mL

Step-by-step explanation:

Answer:

10/9

Step-by-step explanation:

1/3+7/9=10/9

Step-by-step explanation:

= 1 + 7

3 9

= 9+21

27

= 28

27 ans

Answer:

C

Step-by-step explanation:

Answer:

its 61.4x+15

Step-by-step explanation:

Answer:

375/1,000

Step-by-step explanation:

3/8=0.375

375/1,000 =0.375

Answer:31

Step-by-step explanation:

Answer:

b) 31 %

Step-by-step explanation:

Answer:

See explanation

Step-by-step explanation:

Solving (a):

[tex]a_n = 6a_{n-1}[/tex] where [tex]a_0 = 2[/tex]

n = 1

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_1 = 6a_{1-1}[/tex]

[tex]a_1 = 6a_{0}[/tex]

Substitute 2 for [tex]a_0[/tex]

[tex]a_1= 6 * 2[/tex]

[tex]a_1= 12[/tex]

n = 2

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_2 = 6a_{2-1}[/tex]

[tex]a_2 = 6a_{1}[/tex]

Substitute 12 for [tex]a_1[/tex]

[tex]a_2= 6 * 12[/tex]

[tex]a_2= 72[/tex]

n = 3

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_3 = 6a_{3-1}[/tex]

[tex]a_3 = 6a_2[/tex]

Substitute 72 for [tex]a_2[/tex]

[tex]a_3= 6 * 72[/tex]

[tex]a_3= 432[/tex]

n = 4

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_4 = 6a_{4-1}[/tex]

[tex]a_4 = 6a_{3}[/tex]

Substitute 432 for [tex]a_3[/tex]

[tex]a_4 = 6 * 432[/tex]

[tex]a_4 = 2592[/tex]

n = 5

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_5 = 6a_{5-1}[/tex]

[tex]a_5 = 6a_4[/tex]

Substitute 2592 for [tex]a_4[/tex]

[tex]a_5 = 6 * 2592[/tex]

[tex]a_5 = 15552[/tex]

n = 6

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_6 = 6a_{6-1}[/tex]

[tex]a_6 = 6a_{5}[/tex]

Substitute 15552 for [tex]a_5[/tex]

[tex]a_6 = 6 * 15552[/tex]

[tex]a_6 = 93312[/tex]

[tex]a_1= 12[/tex]   [tex]a_2= 72[/tex]   [tex]a_3= 432[/tex]   [tex]a_4 = 2592[/tex]   [tex]a_5 = 15552[/tex]   [tex]a_6 = 93312[/tex]

Solving (b):

[tex]a_n = a^2_{n - 1[/tex] where [tex]a_1 = 2[/tex]

We have

[tex]a_1 = 2[/tex] which serves as the first term

n =2

[tex]a_n = a^2_{n - 1[/tex]

[tex]a_2 = a^2_{2-1}[/tex]

[tex]a_2 = a^2_{1}[/tex]

Substitute 2 for [tex]a_1[/tex]

[tex]a_2 = 2^2[/tex]

[tex]a_2 = 4[/tex]

n = 3

[tex]a_3 = a^2_{3-1}[/tex]

[tex]a_3 = a^2_{2}[/tex]

[tex]a_3 = 4^2[/tex]

[tex]a_3 = 16[/tex]

n = 4

[tex]a_4 = a^2_{4-1}[/tex]

[tex]a_4 = a^2_3[/tex]

[tex]a_4 = 16^2[/tex]

[tex]a_4 = 256[/tex]

n =5

[tex]a_5 = a^2_{5-1[/tex]

[tex]a_5 = a^2_4[/tex]

[tex]a_5 = 256^2[/tex]

[tex]a_5 = 65536[/tex]

n = 6

[tex]a_6 = a^2_{6-1[/tex]

[tex]a_6 = a^2_{5[/tex]

[tex]a_6 = 65536^2[/tex]

[tex]a_6 = 4294967296[/tex]

[tex]a_1 = 2[/tex]   [tex]a_2 = 4[/tex]   [tex]a_3 = 16[/tex]   [tex]a_4 = 256[/tex]   [tex]a_5 = 65536[/tex]   [tex]a_6 = 4294967296[/tex]

Solving (c):

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] [tex]a_0=1[/tex]  ; ;  [tex]a_1=2[/tex]

[tex]a_1=2[/tex] ---- First term

n = 2

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_2=a_{2-1}+3a_{2-2}[/tex]

[tex]a_2=a_1+3a_0[/tex]

Substitute values for a1 and a0

[tex]a_2=2+3 * 1[/tex]

[tex]a_2=2+3[/tex]

[tex]a_2=5[/tex]

n = 3

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_3=a_{3-1}+3a_{3-2}[/tex]

[tex]a_3=a_{2}+3a_{1}[/tex]

[tex]a_3=5+3 * 2[/tex]

[tex]a_3=5+6[/tex]

[tex]a_3=11[/tex]

n = 4

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_4=a_{4-1}+3a_{4-2}[/tex]

[tex]a_4=a_{3}+3a_{2}[/tex]

[tex]a_4=11+3 * 5[/tex]

[tex]a_4=11+15[/tex]

[tex]a_4=26[/tex]

n = 5

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_5=a_{5-1}+3a_{5-2};[/tex]

[tex]a_5=a_{4}+3a_3[/tex]

[tex]a_5=26+3 * 11[/tex]

[tex]a_5=26+33[/tex]

[tex]a_5=59[/tex]

n = 6

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_6=a_{6-1}+3a_{6-2}[/tex]

[tex]a_6=a_{5}+3a_4[/tex]

[tex]a_6=59+3*26[/tex]

[tex]a_6=59+78[/tex]

[tex]a_6=137[/tex]

[tex]a_1=2[/tex]   [tex]a_2=5[/tex]   [tex]a_3=11[/tex]   [tex]a_4=26[/tex]   [tex]a_5=59[/tex]   [tex]a_6=137[/tex]

Solving (d):

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex];     [tex]a_0=1[/tex]; [tex]a_1=1[/tex]

[tex]a_1=1[/tex] --- First term

n = 2

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_2=2 * a_{2-1}+2^2a_{2-2}[/tex]

[tex]a_2=2 * a_1+4*a_0[/tex]

[tex]a_2=2 * 1+4*1[/tex]

[tex]a_2=2 +4[/tex]

[tex]a_2=6[/tex]

n = 3

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_3=3 * a_{3-1}+3^2 * a_{3-2}[/tex]

[tex]a_3=3 * a_{2}+9 * a_{1}[/tex]

[tex]a_3=3 * 6+9 * 1[/tex]

[tex]a_3=18+9[/tex]

[tex]a_3=27[/tex]

n = 4

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_4=4*a_{4-1}+4^2*a_{4-2}[/tex]

[tex]a_4=4*a_{3}+16*a_{2}[/tex]

[tex]a_4=4*27+16*6[/tex]

[tex]a_4=204[/tex]

n = 5

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_5=5 * a_{5-1}+5^2 * a_{5-2}[/tex]

[tex]a_5=5 * a_{4}+25 * a_{3}[/tex]

[tex]a_5=5 * 204+25 *27[/tex]

[tex]a_5=1695[/tex]

n = 6

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_6=6 * a_{6-1}+6^2*a_{6-2}[/tex]

[tex]a_6=6 * a_{5}+36*a_{4}[/tex]

[tex]a_6=6 * 1695+36*204[/tex]

[tex]a_6=17514[/tex]

[tex]a_1=1[/tex]   [tex]a_2=6[/tex]   [tex]a_3=27[/tex]   [tex]a_4=204[/tex]   [tex]a_5=1695[/tex]   [tex]a_6=17514[/tex]

Solving (e):

[tex]a_n= a_{n-1}+a_{n-3};\ a_0=1; a_1=2; a_2=0[/tex]

First term: [tex]a_1=2[/tex]

Second Term: [tex]a_2=0[/tex]

n = 3

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_3= a_{3-1}+a_{3-3}[/tex]

[tex]a_3= a_{2}+a_0[/tex]

[tex]a_3= 0+1[/tex]

[tex]a_3= 1[/tex]

n = 4

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_4= a_{4-1}+a_{4-3}[/tex]

[tex]a_4= a_{3}+a_{1}[/tex]

[tex]a_4= 1+2[/tex]

[tex]a_4=3[/tex]

n = 5

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_5= a_{5-1}+a_{5-3}[/tex]

[tex]a_5= a_{4}+a_{2}[/tex]

[tex]a_5= 3 + 0[/tex]

[tex]a_5= 3[/tex]

n = 6

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_6= a_{6-1}+a_{6-3}[/tex]

[tex]a_6= a_{5}+a_{3}[/tex]

[tex]a_6= 3 + 1[/tex]

[tex]a_6= 4[/tex]

[tex]a_1=2[/tex]   [tex]a_2=0[/tex]   [tex]a_3= 1[/tex]   [tex]a_4=3[/tex]   [tex]a_5= 3[/tex]   [tex]a_6= 4[/tex]

Answer:

it turns out google has the answer

Step-by-step explanation:

can i get brainlyest

Answer:

5/4, 11/12, 1/4, 5/12, 4/10        0.91, 0.89, 0.51, 0.5, 0.04

Step-by-step explanation:

Im pretty sure these are correct

Answer: $8.4

Step-by-step explanation:

To find 30% of 12 first, we move the decimal point to the left two times. Now we have .3 next multiply it by twelve. (3.6) This number is 30% of 12. Lastly, subtract 3.6 from 12 to get your answer.

Answer:

A

Step-by-step explanation:

We are given the equation:  

[tex]x^2+xy-3y=3[/tex]

And we want to find dy/dx at the point (2, 1).

Find dy/dx. We can take the derivative of both sides with respect to x:

[tex]\displaystyle \frac{d}{dx}[x^2+xy-3y]=\frac{d}{dx}[3][/tex]

Implicitly differentiate:

[tex]\displaystyle 2x+y+x\frac{dy}{dx}-3\frac{dy}{dx}=0[/tex]

Solve for dy/dx:

[tex]\displaystyle \begin{aligned} x \frac{dy}{dx} - 3\frac{dy}{dx} & = -2x - y \\ \\ \frac{dy}{dx}(x - 3) & = -2x - y \\ \\ \frac{dy}{dx} & = \frac{-2x-y}{x-3} \\ \\ & = \frac{2x+y}{3-x} \end{aligned}[/tex]

To find dy/dx at (2, 1), evaluate dy/dx for x = 2 and y = 1. Hence:

[tex]\displaystyle \frac{dy}{dx}\Big|_{(2, 1)}=\frac{2(2)+1}{3 - (2)}=\frac{5}{1}=5[/tex]

Hence, our answer is A.

We want to find the differential dy/dx for the given expression at the given point. We will see that the correct option is A.

We have:

x^2 + xy - 3y = 3

First, we need to isolate the variable y in one side of the equation, so we get:

xy - 3y = 3 - x^2

y*(x - 3) = 3 - x^2

y = (x - x^2)/(x - 3)

Now we can differentiate this, remember that if:

f(x) = g(x)*h(x)

then:

f'(x) = g(x)*h'(x) + g'(x)*h'(x)

Using that rule.

dy/dx = (1 - 2x)/(x - 3) - (x - x^2)/(x - 3)^2

Now we need to evaluate this in x = 2 (the x-value of the point) so we get:

(1 - 2*2)/(2 - 3) - (2 - 2^2)/(2 - 3)^2

= (-3)/(-1) - (-2)/(1) = 5

So the correct option is A.

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Answer:

i got D

Step-by-step explanation:

sin 67 = sin 52

21 BC

BC=17.98

BC=18

Answer:

4:14  2:7    8:28

Step-by-step explanation:

The answer fam is........x^3 - 5x^2 + 2x

Answer:

6 units

Step-by-step explanation:

BC = BD/2 = 8

PB² + 8² = 10²

PB² = 100 - 64 = 36 ⇒ PB = 6 units

The length of PB = 6 units

Here, PB is the perpendicular bisector of AC and QC is the perpendicular bisector of BD.

AB = BC = CD

Given that BD = 16 and PC = 10

What is Pythagoras theorem?

Pythagoras theorem is well known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square of the hypotenuse.

Here, BD = 16 and AB = BC = CD

⇒ BC = BD/2 = 16/2 = 8

And, PC = 10

Now using Pythagoras theorem:

[tex]PC^{2} =PB^{2} + BC^{2}[/tex]

[tex]10^{2} =PB^{2} + 8^{2}[/tex]

[tex]PB^{2} + 64 = 100\\PB^{2} = 100 - 64\\PB^{2} = 36\\PB = \sqrt{36}\\PB = 6[/tex]

Hence the length of PB = 6 Units

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Answer:

3 hours

Step-by-step explanation:

20 ÷ 5 = 4 mph

12 ÷ 4 = 3 hours

hope this explanation makes sense! :)

Sara took 3 hours to hike 12 miles if, on the first day of camp, Sara hiked 20 miles in 5 hours and the rate is 4 miles per hour.

Fraction number consists of two parts, one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.

It is given that:

On the first day of camp, Sara hiked 20 miles in 5 hours.

The rate = 20/5 = 4 miles per hour

Let x be the time taken by Sara to hike 12 miles

As its rate of speed of Sara is 4 miles per hour

x = 12/4

The arithmetic operation can be defined as the operation in which we do the addition of numbers, subtraction, multiplication, and division.

x = 3 hours

Thus, Sara took 3 hours to hike 12 miles if, on the first day of camp, Sara hiked 20 miles in 5 hours and the rate is 4 miles per hour.

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Answer:

Negative 1/8 is the answer

Answer:

we dont have the info

Step-by-step explanation:

Answer:

88.9%

Step-by-step explanation:

The result of first step is 6y+3+3y = 5 and values of x and y after solving are 1/4 and 9/2 respectively.

The substitution method is the algebraic method to solve simultaneous linear equations. As the word says, in this method, the value of one variable from one equation is substituted in the other equation.

Given that, the two equations x=6y+3 and x+2y=5 being solved by substitution method

So, the equations are,

x=6y+3.......(i)

x+2y=5......(ii)

Substituting eq(i) in eq(ii)

6y+3+2y = 5

8y = 2

y = 1/4

Put y = 1/4 in eq(i)

x = 6x1/4+3

x = 9/2

Hence, The result of first step is 6y+3+3y = 5 and values of x and y after solving are 1/4 and 9/2 respectively.

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Answer:

[tex]f(x) = -14x + 14[/tex]

Step-by-step explanation:

Given

[tex]Initial\ Value = 14[/tex]

[tex]Rate = -14[/tex]

Required:

Write as a function

Let x be the quantity that changes:

A function is of the form:

[tex]f(x) = mx + b[/tex]

Where

m = slope or rate

b = y intercept or initial value.

So, we can say that:

[tex]m = -14[/tex]

[tex]b = 14[/tex]

Substitute values for b and m in [tex]f(x) = mx + b[/tex]

[tex]f(x) = -14x + 14[/tex]

The answer is 0.7%

Hope this helped!

Answer:

0.6%

Step-by-step explanation:

Percentage = 24/4000 *100

=0.6%