Which of the following could be in section D? A.√99 B.60/15 C.-20 D.9.2

Answer:

Step-by-step explanation:

First equation:

          15 - x = - 3

           -x = - 3 - 15

          -x  =  -18  (now negative signs on b.s cut eachother)

            [x  =  18]

Second equation :

         15 - x = 3

          -x = 3 - 15

          - x = -12 ( now negative sign on b.s cut eachother)

            [x = 12]

so x = 18 and x = 12 are not same

Answer:

23 1/25 desserts per month

Answer:

36, 49

explanation:

The pattern in this series is going by perfect squares:

From this, we can tell the next 2 numbers in the series will be:

Answer:

Step-by-step explanation:

36 divided by 3. I think.

Answer:

Divide 3 by 36 and simplify your answer

Step-by-step explanation:

Answer:

Solving for x:

[tex]x = \frac{3}{2} y + 9[/tex]

Solving for y:

[tex]y = \frac{2}{3} x - 6[/tex]

Step-by-step explanation:

Please see the attached picture for full solution.

Answer

n+9

Step-by-step explanation:

Step-by-step explanation:

35•32 or 1120 there you go

Answer:

Step-by-step explanation:

f(x)= 2x + 5

g(2x + 5)= (2x + 5)^2 = 4x^2 + 20x + 25

h(4x^2 + 20x + 25)= -2(4x^2 + 20x + 25)= -8x^2 -40x - 50

Answer

Step-by-step explanation:

Answer:

5

Step-by-step explanation:

Occurs most often

Answer:

3000000

Step-by-step explanation:

:D

The equivalent value of the equation is A = 3000000

Given data ,

Let the equation be represented as A

Now , the value of A is

Let the first number be p = 3000

Let the second number be q = 1000

Now , A = pq

where A = 3000 x 1000

On simplifying the equation , we get

A = 3000 x 1000

So , the left hand side of the equation is equated to the right hand side by the value of ( 3000 x 1000 )

A = 3000000

Therefore , the value of A = 3000000

Hence , the expression is A = 3000000

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Answer:

1/2

Step-by-step explanation:

Answer:

[tex]\frac{1}{2}[/tex]

Step-by-step explanation:

28 * 2 = 56.

You selected the correct answer choice.

The given equation y = 3x-5 is in the form y = mx+b with slope m = 3.

Parallel lines have equal slopes but different y intercepts. So any parallel line will also have slope m = 3 (the b value must be different though).

Answer:

X=7 ST=11 RT=17

Step-by-step explanation:

RT=RS+ST. RS= 2(7)-8. ST=11

X+10=2x-8+11. =14-8=6. RT= x+10

X+10=2x+3. RS=6. 7+10=17

10=x+3. RT=17

X=7

Answer:

32

Step-by-step explanation:

You can keep adding 4 from -16 for a total of 12 times to get 32

or you can multiply 12*4 to get 48 and subtract 16 from it

hope this is good enough

Answer:

Step-by-step explanation:

83/100 is already simplified. You can not make it smaller.

Answer:

The Communitive Property of Addition

Step-by-step explanation:

The properties are

Communitive: You can switch the order and still get the same answer

Assosiative: You can switch what numbers are in the parentheses and still get the same answer (if you're only adding) a+(b+c)=c+(a+b)

Distributive: (a*b)+(a*c)=a*(b+c)

Identity: You can add a number with 0 and get the same number you started with

y+x=x+y is the communitive property because, no matter which order the numbers are, you get the same answer

Answer:

5/3

Step-by-step explanation:

1.6 repeating is equal to 5/3

The repeating decimal 1.6 can be written as the simplified fraction 5/3.

To rewrite the decimal 1.6 repeating (1.6) as a simplified fraction, we can use algebraic manipulation. Let's denote 1.6 as x:

x = 1.6

Now, to eliminate the repeating decimal, we can multiply both sides of the equation by 10 (since there is one digit repeating after the decimal point):

10x = 16.6

Next, subtract the original equation from the multiplied equation to eliminate the repeating decimal:

10x - x = 16.6 - 1.6

This simplifies to:

9x = 15

Finally, divide both sides by 9 to solve for x:

x = 15 / 9

Now, we simplify the fraction:

x = 5/3

Therefore, the repeating decimal 1.6 can be written as the simplified fraction 5/3.

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Answer:

200+10+6=216

Step-by-step explanation:

2 hundred and 16 is in word form

Since 200 can be written as a sum of 2 100's, hence there are two hundred in the expression given,

Given the expression 200 + 10 + 6

Each term can be written in words as shown:

200 is two hundred

200 = 100 + 100

Since 200 can be written as a sum of 2 100's, hence there two hundred in the expression given,

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Answer:

x = -18

Step-by-step explanation:

4+ x / -9 = 6

x / -9 = 6 - 4

x / -9 = 2

x = 2*-9

x = -18

found this deimos Google hope this helps -

x in (-oo:+oo)

x/4-9 = 6 // - 6

x/4-9-6 = 0

1/4*x-15 = 0 // + 15

1/4*x = 15 // : 1/4

x = 15/1/4

x = 60

Answer:

21.6 Pounds

Step-by-step explanation:

Multiply 2.5 by 8.34 to get 20.85 pounds of water

add the weight of the bucket (.75) to the weight of the water (20.85) to get 21.6 pounds total

Answer:

Step-by-step explanation:

x  - the miles to travel

467+x  - the main odometer after x miles travel

22+x   - the trip odometer after x miles travel

2(22+x)    - two times the trip odometer after x miles travel

2(22+x) = 467 + x

44 + 2x = 467 + x

     - 44       -44

     2x = 423 + x

    -x       -x

     x = 423

Answer:

how far you've driven

Step-by-step explanation:

Answer:

2 between 2 and 4 and 8 between 7 and -1

Answer: C. -10, A -6, B 2

Step-by-step explanation:

Answer:

f(-5) = 50

Step-by-step explanation:

Simply plug in -5 for x into the function:

f(-5) = (-5)² - 5(-5)

f(-5) = 25 + 25

f(-5) = 50

Since , angle PQT and PQR are in a line .

So ,  PQT=PQR

(3x + 47) + (6x - 25) = 180

x = 17.56

Now , PQR and SQR are also in same line .

So , PQR + SQR = 180

( 6x - 25 ) + SQR = 180

105.33 - 25 + SQR = 180

SQR = 99.67°

Hence , this is the required solution .

Answer: - 1°F

Step-by-step explanation:

Given that:

At 8a.m:

Temperature = 3°F below 0

By 1p.m ; Temperature rose by 14F ;

By 10pm ; temperature dropped by 12F

Temperature at 10 pm:

3°F below 0 = - 3°F

Temperature at 1pm: temperature increased by 14°F

-3°F + 14°F = 11°F

Temperature at 10pm: temperature dropped by 12°F

11°F - 12 = - 1°F

Answer:

C) 1/2

Step-by-step explanation:

The slope of a perpendicular line is the negative inverse.

First, convert the given line to slope-intercept form.

2x + y = -7

y = -2x − 7

The slope of this line is -2.  So the slope of the perpendicular line is:

m = -1 / (-2)

m = 1/2

Answer:

-8 and 4 are 6 away from -2

Step-by-step explanation:

------/----------------------------------------/----

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

Answer:

Part A: The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B: For the total distance is as close to 3 miles as possible, the start point of the parade should be at the point on Broadway with coordinates (9.941, 4.970)

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue; (4, 2)

For Broadway; (7.97, 2.49)

Step-by-step explanation:

Part A: The length of the given route can be found using the equation for the distance, l, between coordinate points as follows;

[tex]l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}[/tex]

Where for the Broadway potion of the parade route, we have;

(x₁, y₁) = (12, 3)

(x₂, y₂) = (6, 0)

[tex]l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}[/tex]

For the Central Avenue potion of the parade route, we have;

(x₁, y₁) = (6, 0)

(x₂, y₂) = (2, 4)

[tex]l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}[/tex]

Therefore, the total length of the parade route =-3·√5 + 4·√2 = 12.265 unit

The scale of the drawing is 1 unit = 0.25 miles

Therefore;

The actual length of the initial parade =0.25×12.265 unit = 3.09 miles

The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B:

For an actual length of 3 miles, the length on the scale drawing should be given as follows;

1 unit = 0.25 miles

0.25 miles = 1 unit

1 mile =  1 unit/(0.25) = 4 units

3 miles = 3 × 4 units = 12 units

With the same end point and route, we have;

[tex]l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}[/tex]

y² + (6 - x)² = 176 - 96·√2

y² = 176 - 96·√2 - (6 - x)²............(1)

Also, the gradient of l₁ = (3 - 0)/(12 - 6) = 1/2

Which gives;

y/x = 1/2

y = x/2 ..............................(2)

Equating equation (1) to (2) gives;

176 - 96·√2 - (6 - x)² = (x/2)²

176 - 96·√2 - (6 - x)² - (x/2)²= 0

176 - 96·√2 - (1.25·x²- 12·x+36) = 0

Solving using a graphing calculator, gives;

(x - 9.941)(x + 0.341) = 0

Therefore;

x ≈ 9.941 or x = -0.341

Since l₁ is required to be 12 - 4·√2, we have and positive, we have;

x ≈ 9.941 and y = x/2 ≈ 9.941/2 = 4.97

Therefore, the start point of the parade should be the point (9.941, 4.970) on Broadway so that the total distance is as close to 3 miles as possible

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue;

Camera location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)

For Broadway;

Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).

This exercise is necessary to use the given map information and then make the distance between points, in this way we find that:

A)The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles.

B) Broadway and Cedar Street

C) [tex]Central \ Avenue(4, 2)Broadway (7.97, 2.49)[/tex]

So from the distance between points and the informed map we have:

A) The equation for the distance, as follows:

[tex]l=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]

Where for the Broadway potion of the parade route, we have:

[tex](x_1,y_1)=(12,3)\\(x_2,y_2)=(6,0)[/tex]

Applying the values ​​in the formula given above, we have:

[tex]l_1=\sqrt{(0-3)^2+(6-12)^2}=3\sqrt{5}[/tex]

For the Central Avenue potion of the parade route, we have:

[tex](x_1, y_1) = (6, 0)(x_2, y_2) = (2, 4)[/tex]

Applying the values ​​in the formula given above, we have:

[tex]l_2=\sqrt{(4-0)^2+(2-6)^2}=4\sqrt{2}[/tex]

Therefore, the total length of the parade route:

[tex]3\sqrt{5} + 4\sqrt{2} = 12.265 \\0.25*12.265 unit = 3.09\ miles[/tex]

B) The two streets that are the same distance apart are the streets: Broadway and Cedar Street.

C) The coordinates of the cameras stationed half way down each road are:

For Central Avenue:

[tex]Camera \ location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)[/tex]

For Broadway:

[tex]Camera\ location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49)[/tex]

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