Model the situation.

The sum of two consecutive integers is 85. Find the integers that satisfy this condition.
Answer choices

A) 1 + 2 = 85

B) x+(x+1)=85

C) x+x=85

D) x+2x=85

9514 1404 393

Answer:

  10, -1

Step-by-step explanation:

If you assume the numbers are integers, you can look at the factor pairs that make -10:

  -10 = 10(-1) = 5(-2) = 2(-5) = 1(-10)

The sums of these pairs are 9, 3, -3, -9, so the pair of interest is the first one:

  10, -1

Answer:

x = 7

Step-by-step explanation:

[tex]8x-3=11+6x\\\\2x-3=11\\\\2x=14\\\\x=7[/tex]

Answer:

x = 7

Step-by-step explanation:

8x - 3 = 11 + 6x

8x - 3 + 3 = 11 + 6x +3

8x = 6x + 14

8x - 6x = 6x + 14 - 6x

2x = 14

2x/2 = 14/2

x = 7

Answer:

(-1,-5)

Step-by-step explanation:

To understand where this point was before the transformation you must first understand the transformation itself. Shifting left is a horizontal translation, this means that it will affect the x-coordinate. The x-coordinate is the first digit in the pair. Therefore, in the case of (3,5), 3 would be the x-coordinate. Additionally, moving left is moving towards the negative, so it is the same as subtracting. Thus, to find where you started, work backward. Do this by adding 4 to the x-coordinate. This means the beginning coordinate was (-1,-5).

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given function:

This is a quadratic function with positive leading coefficient, therefore it is an increasing function with its minimum at vertex. It has no maximum value.

The vertex is at x = -b/2a:

The minimum value is:

The range of the function is:

f(x)=x^2+8x+8

[tex]\\ \sf\longmapsto \dfrac{-b}{2a}[/tex]

[tex]\\ \sf\longmapsto \dfrac{-8}{2(1)}[/tex]

[tex]\\ \sf\longmapsto \dfrac{-8}{2}[/tex]

[tex]\\ \sf\longmapsto -4[/tex]

[tex]\\ \sf\longmapsto f(-4)[/tex]

[tex]\\ \sf\longmapsto (-4)^2+8(-4)+8[/tex]

[tex]\\ \sf\longmapsto 16-32+8[/tex]

[tex]\\ \sf\longmapsto -16+8[/tex]

[tex]\\ \sf\longmapsto -8[/tex]

The range is

[tex]\\ \sf\longmapsto (8,\infty)[/tex]

Answer: 7

Step-by-step explanation:

1. multiply both sides by -6 so the -6 on the right side could cancel out and the left side would be -6:  1= a-13/-6 ==> -6=a-13

2. add both sides by 13 so -13 on the right side could cancel out and the left side would be -6+13: -6=a-13 ==> -6+13=a-13+13

3. you know -6+13 equals 7 right? so the answer is a=7

Hey there! I'm happy to help!

We want to get the a all by itself. For me, I always like to have the variable on the left side, so first, let's just flip this around.

[tex]\frac{a-13}{-6} =1[/tex]

What we want to do is get all of the constants (numbers without variables) on the right side. This will leave a all on its own.

We see that the quantity a-13 is being divided by -6. To undo this, we will just multiply both sides by -6. This will cancel out that -6 and get us closer to having the a isolated.

[tex]\frac{a-13}{-6} *-6=1*-6\\\frac{a-13}{1}=-6\\a-13=-6[/tex]

Now we add 13 to both sides to undo the -13 on the left side.

[tex]a-13+13=-6+13[/tex]

[tex]a+0=7\\a=7[/tex]

Have a wonderful day and keep on learning! :D

Step-by-step explanation:

we can also write them as,

Now, from -a²-2ab-b² if we take - as common term then we get,

Now, by using the concept of a²+2ab+b²=(a+b)²

I hope it helped U

stay safe stay happy

Answer:

1/4= 0.5=0.125, 2/4=0.5=0.125, 3/4= 0.75=0.1878

Answer:

2

Step-by-step explanation:

Answer:

230cm²

Step-by-step explanation:

We can split the shape into 2 parts - the triangle and the rectangle.

The area of the retangle is 10*17cm = 170cm²

The area of the triangle  = 1/2 * (22-10) * (17-7)

                                         =  1/2 * 12 * 10 = 60

Therefore the total area is equal to 170+60 = 230cm²

Answer:

Decimal Fraction Percentage

0.101 101/1000 10.1%

0.1 100/1000 10%

0.1013 101/997 10.13%

0.1012 101/998 10.12%

Step-by-step explanation:

The critical points of h(x,y) occur wherever its partial derivatives [tex]h_x[/tex] and [tex]h_y[/tex] vanish simultaneously. We have

[tex]h_x = 8-4y-8x = 0 \implies y=2-2x \\\\ h_y = 10-4x-12y^2 = 0 \implies 2x+6y^2=5[/tex]

Substitute y in the second equation and solve for x, then for y :

[tex]2x+6(2-2x)^2=5 \\\\ 24x^2-46x+19=0 \\\\ \implies x=\dfrac{23\pm\sqrt{73}}{24}\text{ and }y=\dfrac{1\mp\sqrt{73}}{12}[/tex]

This is to say there are two critical points,

[tex](x,y)=\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\text{ and }(x,y)=\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)[/tex]

To classify these critical points, we carry out the second partial derivative test. h(x,y) has Hessian

[tex]H(x,y) = \begin{bmatrix}h_{xx}&h_{xy}\\h_{yx}&h_{yy}\end{bmatrix} = \begin{bmatrix}-8&-4\\-4&-24y\end{bmatrix}[/tex]

whose determinant is [tex]192y-16[/tex]. Now,

• if the Hessian determinant is negative at a given critical point, then you have a saddle point

• if both the determinant and [tex]h_{xx}[/tex] are positive at the point, then it's a local minimum

• if the determinant is positive and [tex]h_{xx}[/tex] is negative, then it's a local maximum

• otherwise the test fails

We have

[tex]\det\left(H\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\right) = -16\sqrt{73} < 0[/tex]

while

[tex]\det\left(H\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)\right) = 16\sqrt{73}>0 \\\\ \text{ and } \\\\ h_{xx}\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-8 < 0[/tex]

So, we end up with

[tex]h\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-\dfrac{4247+37\sqrt{73}}{72} \text{ (saddle point)}\\\\\text{ and }\\\\h\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)=-\dfrac{4247-37\sqrt{73}}{72} \text{ (local max)}[/tex]

Answer:

Step-by-step explanation:

[tex]\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}[/tex]

Step-by-step explanation:

steps are in picture above.

if you need to ask questions I'll give you answers gladly.

Answer:

3x-721/44 over 7

Step-by-step explanation:

1)Combine multiplied terms into a single fraction

2)Multiply the numbers

3)Subtract the numbers

4)Find common denominator

5)Combine fractions with common denominator

6)Multiply the numbers

7)Rearrange terms

9514 1404 393

Answer:

Step-by-step explanation:

Let b and c represent the rates of the boat and current, respectively. Then the upstream rate is ...

  speed = distance/time

  b -c = (116 km)/(4 h) = 29 km/h

  b +c = (164 km)/(4 h) = 41 km/h

Adding these equations gives ...

  2b = 70 km/h

  b = 35 km/h

Substituting into the first equation, we have ...

  35 -c = 29

  c = 6

Rate of the boat: 35 km/h.

Rate of the current: 6 km/h

Step-by-step explanation:

did you give us all the information ? the equating looks cut off.

so, from what I can see we have

9(y-4) - 3y = 2(3y-2)

let's try :

9y - 36 - 3y = 6y - 4

0 = 32

so, no, this can't be it. this does not have a solution. you must have cut off some additional information.

I think the answer you are looking for is 7n, which also means 7 times n.

Answer:

7x

Step-by-step explanation:

Product refers to multiplication. "A number" refers to an unknown number otherwise known as a variable. In this case, I use x.

Answer:

place value of 2 is 100

Step-by-step explanation:

please brainleist my answer

Answer:

The place value of 2 in 6.253 is 200

Step-by-step explanation:

Place value of 2 is occurring in hundreds

Answer:

z = 2y - 4x

Step-by-step explanation:

[tex]{ \sf{x = \frac{2y - z}{4} }} [/tex]

multiply both sides by 4:

[tex]{ \sf{(4 \times x) = ( \frac{2y - z}{4}) \times 4 }} \\ \\ { \sf{4x = 2y - z}} \\{ \sf{z = 2y - 4x}}[/tex]

24 - (5 - 3)⁴ = 24 - 2⁴ = 24 - 16 = 8

8 : 4 = 2 ← the end solution

1.

Paul's present age = x

John's present age = 18

Paul's age 4 years ago = x - 4

John's age 4 years ago = 18 - 4 = 14

Also,

Paul's age four years ago was twice that of John's age four years ago.

Therefore,

Paul's age 4 years ago = 2(14)

= 28

Paul's age 4 years ago = x - 4 = 28

x - 4 = 28

x = 28 + 4

x = 32

Therefore, the present age of Paul is 32.

2.

Cost of one bottle of milk = $3.2

Cost of five bottles of milk = $3.2 × 5 = $15

Cost of four bottles of juice = $15 - $4.4

= $10.6

Cost of one bottle of juice = x

= Cost of four bottles of juice

4

= 10.6

4

= 2.56

Therefore, one bottle of juice costs $2.56.

3.

Savings spent on personal ornaments = 1/8

Savings spent on clothing = 5/6

Total savings spent = 1/8 + 5/6

= 3/24 + 20/24

= (3+20)/24

= 23/24

Savings left = 1/24 = $130

Total amount of savings = x

As 130 is one part of the total savings,

Total amount of savings =

x = 130 × 24

x = 3120

Therefore, original amount of savings is $3120.

Answer:

126720

Step-by-step explanation:

Step-by-step explanation:

Janyce's rule doesn't work. In the case where n = 3 her rule predicts that there will be 3(3) - 1 = 8 squares but instead, there are 10 squares. Likewise, for n = 4, there are 17 squares instead of 11 that she predicted. Instead, I propose the rule [tex]n^2 + 1.[/tex] It works for all the figures shown and mostly likely, for all other values of n.

2. Part A;

Given that the number of squares obtained using Janyce conjecture for

figures 3 and 4 which are 8 and 11 squares respectively and the number of

squares in the diagram for figures 3 and 4 which are 10 and 17 squares

respectively are not the same, disproves Janyce conjecture

Part B:

The rule for the number of squares in the nth squares is n² + 1

The reasons the above conclusion and rule are correct is as follows:

2. Part A

The given number of squares in each figure are;

[tex]\begin{array}{|c|c|c|}\mathbf{Figure \ number}&\mathbf{Number \ of \ squares} &\mathbf{Janyce \ conjecture \ (3\cdot n - 1)}\\&&\\1&2 \ squares &(3 \times 1 - 1 = 2) \ squares \\2&5 \ squares&(3 \times 2 - 1 = 5) \ squares\\3&10 \ squares &(3 \times 3 - 1 = 8) \ squares\\4&17 \ squares&(3 \times 4 - 1 = 11) \ squares\end{array}[/tex]

Given that the number squares in figure 3, and 4 are;

Figure 3 = 10 squares

Figure 4 = 17 squares

Janyce conjectured number of squares are;

Figure 3 = 8 squares

Figure 4 = 11 squares

The values for the number of squares in the diagram in figures 3 and 4 are different from Janyce conjecture which disproves Janyce's conjecture for the number of squares in the nth figure

Part B

From the diagrams, we have;

The number of squares on the each figure is given by a square shape with a side length equal to the figure number of squares plus one extra square at the top right end corner, we have;

The area of the square part of the shape = n² = The number of squares in the square shape

The number of squares in the nth figure = n² + 1

Verifying, we have;

Figure 1 = (1² + 1) squares = 2 squares (Same as in diagram)

Figure 2 = (2² + 1) squares = 5 squares (Same as in diagram)

Figure 3 = (3² + 1) squares = 10 squares (Same as in diagram)

Figure 4 = (4² + 1) squares = 17 squares (Same as in diagram)

Therefore;

The correct formula for the number of squares in the nth figure = n² + 1

Learn more about the nth term of a sequence here:

https://brainly.com/question/12998719

https://brainly.com/question/15325765

https://brainly.com/question/21493252

Answer:

[-5, 2]

Step-by-step explanation:

We have to find the interval for which f(x) <= 0, the interval is [-5, 2].

Answer:

(0,-5)

Step-by-step explanation:

4x-2(-2x-5)+10

x=0

y=-2*0-5

y=-5

Answer:

45000rupees

Step-by-step explanation:

cost of 1 litre of petrol=60rupees

cost of 750 litres of petrol=750×60rupees

=45000rupees

Let D denote the region whose boundary is C. By Green's theorem,

[tex]\displaystyle \int_C 8y^2x\,\mathrm dx + 6x^2y\,\mathrm dy = \iint_D \frac{\partial(6x^2y)}{\partial x} - \frac{\partial(8y^2x)}{\partial y} \,\mathrm dx\,\mathrm dy \\\\ = \int_0^3 \int_0^3 (12xy-16yx)\,\mathrm dx\,\mathrm dy \\\\ = -4 \int_0^3 \int_0^3 xy\,\mathrm dx\,\mathrm dy = \boxed{-81}[/tex]