Are these quanities dirctly proportional?
Distance and Time it takes to cover it, if a car moves with the same speed.
Plz Help Me
Answer Yes or No

Answer:

look below.

Step-by-step explanation:

as z= 7

3z -3

3(7) -3 =21-3 =18

The ratio of the number of eggs in three boxes to the total number of eggs is 1: 15 in its simplest terms.

Suppose the real measurements were "a" and "b". Then their ratio will be formed as;

[tex]\dfrac{a}{b} = \dfrac{p \times x}{q \times x} = \dfrac{p}{q}[/tex]

where is a common factor (not 1)? This shows that to get the real measurements from the given ratio, we need to assume to have some factor possibly canceled from both the numerator and the denominator.

Thus, a = px, b = qx

We have given that farmer 30 boxes of eggs. and there are 6 eggs in each box.

Thus, we know that the total number of eggs is the same as the number of eggs given in 30 boxes.

Hence, the ratio will be;

2: 30,

then divide both sides by 2 ;

1: 15

Therefore, the ratio of the number of eggs in three boxes to the total number of eggs is 1: 15 in its simplest terms.

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Answer:

Step-by-step explanation:

t22 = a + (n- 1)d

t22 = 3 + 21*4

t22 = 3 + 84

t22 = 87

Answer:

I think its C (I think..im doing the test so ill tell u if its right or wrong)

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

Blake should not include the outlier because it overstates the effect of rainfall on temperature.

Answer:

900 cubic inches.

Step-by-step explanation:

Given the following data

Volume of right circular cone = 300in³

We know that the volume of a right circular cone is given by the formula;

[tex] V = \frac{1}{3} \pi r^{2}h[/tex]

Where;

The volume of a right circular cylinder is given by the formula;

[tex] V = \pi r^{2}h[/tex]

Thus, multiplying the volume of the right circular cone by 3 would give us the volume of the right circular cylinder.

[tex] V = \frac{1}{3} \pi r^{2}h * 3[/tex]

Substituting into the equation, we have;

V = 300 * 3

V = 900in³

Therefore, the volume of a right cylinder that has the same base and height as the cone is 900 cubic inches.

Answer:

Step-by-step explanation:

Given that :

Dimension of rail road toy trains :

Length = 15 inches

Width = 4 inches

Scale = 2/9

Dimension of car on the long Island rail road :

Let length = x

Length = 2/9 of x = 15 inches

= 0.22222x = 15 inches

LENGTH = 67.5 inches

WIDTH = 2/9 of x = 4 inches

0.22222x = 4 inches

Width = 18

Width = 6 inches

Answer:

hi

Step-by-step explanation:

Answers:

===================================================

Explanation:

The largest exponent term is what will go first. Then the term with the second largest goes next, and so on. We count the exponents down. The exponents will count down in this case as 5, 3, 1, 0

Note: -9 = -9*1 = -9m^0 and -11m = -11m^1

So that's why standard form is [tex]\frac{3}{5}m^5 + 7m^3 - 11m-9[/tex]

We can write this as (3/5)m^5 + 7m^3 - 11m - 9

------------------------

The leading term is the first term after getting everything in standard form. The coefficient of the leading term is the leading coefficient. So in this case, it would be 3/5.

The degree is the largest exponent. So the degree is 5. This is a quintic polynomial.

Answer:

A

Step-by-step explanation:

Answer:

m(x) = 5 + 100x

Step-by-step explanation:

Let m(x) be the balance of the account after x months.

m(x) = 5 + 100x

Answer:

5/14

Step-by-step explanation:

5 + 9 = 14- so our denominator is 14

Theres 5 milk chocolates, so the probability is 5/14

I hope this helped, please mark Brainliest, thank you !!

No, because intergers don't include fractions.

Answer:

No

Step-by-step explanation:

Integers cannot have any fractions.

If you simplify this to get 8/1 and then 8, 8 is an integer

Answer:

( -8 , 3 )  8 Down 3 Across

Answer:

The solution is too long. So, I included them in the explanation

Step-by-step explanation:

This question has missing details. However, I've corrected each question before solving them

Required: Determine the inverse

1:

[tex]f(x) = 25x - 18[/tex]

Replace f(x) with y

[tex]y = 25x - 18[/tex]

Swap y & x

[tex]x = 25y - 18[/tex]

[tex]x + 18 = 25y - 18 + 18[/tex]

[tex]x + 18 = 25y[/tex]

Divide through by 25

[tex]\frac{x + 18}{25} = y[/tex]

[tex]y = \frac{x + 18}{25}[/tex]

Replace y with f'(x)

[tex]f'(x) = \frac{x + 18}{25}[/tex]

2. [tex]g(x) = \frac{12x - 1}{7}[/tex]

Replace g(x) with y

[tex]y = \frac{12x - 1}{7}[/tex]

Swap y & x

[tex]x = \frac{12y - 1}{7}[/tex]

[tex]7x = 12y - 1[/tex]

Add 1 to both sides

[tex]7x +1 = 12y - 1 + 1[/tex]

[tex]7x +1 = 12y[/tex]

Make y the subject

[tex]y = \frac{7x + 1}{12}[/tex]

[tex]g'(x) = \frac{7x + 1}{12}[/tex]

3: [tex]h(x) = -\frac{9x}{4} - \frac{1}{3}[/tex]

Replace h(x) with y

[tex]y = -\frac{9x}{4} - \frac{1}{3}[/tex]

Swap y & x

[tex]x = -\frac{9y}{4} - \frac{1}{3}[/tex]

Add [tex]\frac{1}{3}[/tex] to both sides

[tex]x + \frac{1}{3}= -\frac{9y}{4} - \frac{1}{3} + \frac{1}{3}[/tex]

[tex]x + \frac{1}{3}= -\frac{9y}{4}[/tex]

Multiply through by -4

[tex]-4(x + \frac{1}{3})= -4(-\frac{9y}{4})[/tex]

[tex]-4x - \frac{4}{3}= 9y[/tex]

Divide through by 9

[tex](-4x - \frac{4}{3})/9= y[/tex]

[tex]-4x * \frac{1}{9} - \frac{4}{3} * \frac{1}{9} = y[/tex]

[tex]\frac{-4x}{9} - \frac{4}{27}= y[/tex]

[tex]y = \frac{-4x}{9} - \frac{4}{27}[/tex]

[tex]h'(x) = \frac{-4x}{9} - \frac{4}{27}[/tex]

4:

[tex]f(x) = x^9[/tex]

Replace f(x) with y

[tex]y = x^9[/tex]

Swap y with x

[tex]x = y^9[/tex]

Take 9th root

[tex]x^{\frac{1}{9}} = y[/tex]

[tex]y = x^{\frac{1}{9}}[/tex]

Replace y with f'(x)

[tex]f'(x) = x^{\frac{1}{9}}[/tex]

5:

[tex]f(a) = a^3 + 8[/tex]

Replace f(a) with y

[tex]y = a^3 + 8[/tex]

Swap a with y

[tex]a = y^3 + 8[/tex]

Subtract 8

[tex]a - 8 = y^3 + 8 - 8[/tex]

[tex]a - 8 = y^3[/tex]

Take cube root

[tex]\sqrt[3]{a-8} = y[/tex]

[tex]y = \sqrt[3]{a-8}[/tex]

Replace y with f'(a)

[tex]f'(a) = \sqrt[3]{a-8}[/tex]

6:

[tex]g(a) = a^2 + 8a- 7[/tex]

Replace g(a) with y

[tex]y = a^2 + 8a - 7[/tex]

Swap positions of y and a

[tex]a = y^2 + 8y - 7[/tex]

[tex]y^2 + 8y - 7 - a = 0[/tex]

Solve using quadratic formula:

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]a = 1[/tex] ; [tex]b = 8[/tex]; [tex]c = -7 - a[/tex]

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex] becomes

[tex]y = \frac{-8 \±\sqrt{8^2 - 4 * 1 * (-7-a)}}{2 * 1}[/tex]

[tex]y = \frac{-8 \±\sqrt{64 + 28 + 4a}}{2 * 1}[/tex]

[tex]y = \frac{-8 \±\sqrt{92 + 4a}}{2 * 1}[/tex]

[tex]y = \frac{-8 \±\sqrt{92 + 4a}}{2 }[/tex]

Factorize

[tex]y = \frac{-8 \±\sqrt{4(23 + a)}}{2 }[/tex]

[tex]y = \frac{-8 \±2\sqrt{(23 + a)}}{2 }[/tex]

[tex]y = -4 \±\sqrt{(23 + a)}[/tex]

[tex]g'(a) = -4 \±\sqrt{(23 + a)}[/tex]

7:

[tex]f(b) = (b + 6)(b - 2)[/tex]

Replace f(b) with y

[tex]y = (b + 6)(b - 2)[/tex]

Swap y and b

[tex]b = (y + 6)(y - 2)[/tex]

Open Brackets

[tex]b = y^2 + 6y - 2y - 12[/tex]

[tex]b = y^2 + 4y - 12[/tex]

[tex]y^2 + 4y - 12 - b = 0[/tex]

Solve using quadratic formula:

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]a = 1[/tex] ; [tex]b = 4[/tex]; [tex]c = -12 - b[/tex]

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex] becomes

[tex]y = \frac{-4\±\sqrt{4^2 - 4 * 1 * (-12-b)}}{2*1}[/tex]

[tex]y = \frac{-4\±\sqrt{4^2 - 4 *(-12-b)}}{2}[/tex]

Factorize:

[tex]y = \frac{-4\±\sqrt{4(4 - (-12-b))}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{(4 - (-12-b))}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{(4 +12+b)}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{16+b}}{2}[/tex]

[tex]y = -2\±\sqrt{16+b}[/tex]

Replace y with f'(b)

[tex]f'(b) = -2\±\sqrt{16+b}[/tex]

8:

[tex]h(x) = \frac{2x+17}{3x+1}[/tex]

Replace h(x) with y

[tex]y = \frac{2x+17}{3x+1}[/tex]

Swap x and y

[tex]x = \frac{2y+17}{3y+1}[/tex]

Cross Multiply

[tex](3y + 1)x = 2y + 17[/tex]

[tex]3yx + x = 2y + 17[/tex]

Subtract x from both sides:

[tex]3yx + x -x= 2y + 17-x[/tex]

[tex]3yx = 2y + 17-x[/tex]

Subtract 2y from both sides

[tex]3yx-2y =17-x[/tex]

Factorize:

[tex]y(3x-2) =17-x[/tex]

Make y the subject

[tex]y = \frac{17 - x}{3x - 2}[/tex]

Replace y with h'(x)

[tex]h'(x) = \frac{17 - x}{3x - 2}[/tex]

9:

[tex]h(c) = \sqrt{2c + 2}[/tex]

Replace h(c) with y

[tex]y = \sqrt{2c + 2}[/tex]

Swap positions of y and c

[tex]c = \sqrt{2y + 2}[/tex]

Square both sides

[tex]c^2 = 2y + 2[/tex]

Subtract 2 from both sides

[tex]c^2 - 2= 2y[/tex]

Make y the subject

[tex]y = \frac{c^2 - 2}{2}[/tex]

[tex]h'(c) = \frac{c^2 - 2}{2}[/tex]

10:

[tex]f(x) = \frac{x + 10}{9x - 1}[/tex]

Replace f(x) with y

[tex]y = \frac{x + 10}{9x - 1}[/tex]

Swap positions of x and y

[tex]x = \frac{y + 10}{9y - 1}[/tex]

Cross Multiply

[tex]x(9y - 1) = y + 10[/tex]

[tex]9xy - x = y + 10[/tex]

Subtract y from both sides

[tex]9xy - y - x = y - y+ 10[/tex]

[tex]9xy - y - x = 10[/tex]

Add x to both sides

[tex]9xy - y - x + x= 10 + x[/tex]

[tex]9xy - y = 10 + x[/tex]

Factorize

[tex]y(9x - 1) = 10 + x[/tex]

Make y the subject

[tex]y = \frac{10 + x}{9x - 1}[/tex]

Replace y with f'(x)

[tex]f'(x) = \frac{10 + x}{9x -1}[/tex]

[tex]\text {Hello! Let's Solve this Problem!}[/tex]

[tex]\text {\underline {The First Step is to Add 4}}[/tex]

[tex]\text {3x-4+4=14+4}[/tex]

[tex]\text {3x=18}[/tex]

[tex]\text {\underline {The Final Step is to Divide 3}}[/tex]

[tex]\text {3x/3=18/3}[/tex]

[tex]\text {Your Answer Would Be:}[/tex]

[tex]\Huge\boxed {x=6}[/tex]

[tex]\text {Best of Luck!}[/tex]

[tex]\huge\text{$x=\boxed{6}$}[/tex]

We can solve for the variable [tex]x[/tex] by isolating it on one side of the equals sign.

Start by adding [tex]4[/tex] to both sides to cancel out the [tex]-4[/tex] on the left side of the equation.

[tex]\begin{aligned}3x-4+4&=14+4\\3x&=18\end{aligned}[/tex]

Now divide both sides of the equation by [tex]3[/tex] since [tex]3x[/tex] is the same as [tex]3*x[/tex] and we want to isolate [tex]x[/tex], keeping in mind that anything multiplied by [tex]1[/tex] is itself.

[tex]\begin{aligned}3x&=18\\3*x&=18\\3*x\div3&=18\div3\\x*3\div3&=18\div3\\x*1&=6\\x&=\boxed{6}\end{aligned}[/tex]

Answer:

Hi it’s me

Step-by-step explanation:

7is the first one

10 is the second

Nvm you already got the answer cause yo asked this 2 weeks ago anyways hi

Answer:

x < 1

Step by step explanation:

10 + 4x < 14

4x < 14 -10

4x < 4

x < 1

When you graph it on a number line, draw a line that extends from 1 towards the left side (since x is less that 1) and use unshaded circles.

Answer:

x = by + 3b

Step-by-step explanation:

I hope this helps ya!

The difference between consecutive first and second integers is increased by 1 thus the next number 19 - x = 5 → x = 14.

A pattern is a specific arrangement of numbers or any term which will follow a specific code or hint.

For example 2,4,6,8 in this, the difference between any two-term is only 2 and this will be its specific pattern.

As per the given pattern,29, 28, 26, 23, 19

29 - 28 = 1

28 - 26 = 2

26 - 23 = 3

23 - 19 = 4

Since the difference is increasing by 1

So the difference of 19 and the next number (x) will be 4 + 1 = 5.

19 - x = 5

x = 14

Hence "The difference between consecutive first and second integers is increased by 1 thus the next number 19 - x = 5 → x = 14".

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Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

21

Step-by-step explanation:

140*0.15 is 21

Answer:

21

Step-by-step explanation:

Answer:

x = -2

Step-by-step explanation:

Answer: 600

Step-by-step explanation:

I just took the test

Answer:

A ( SAS )

Step-by-step explanation:

Answer:

The most that they can spend in food is $239

Step-by-step explanation:

Their budget allows for less than $720 for both food and rent.

We are told that this budget allows

twice as much money for rent as

for food.

Thus,

Let x be amount of spent on food.

Thus, money spent on rent is 2x

Thus,

2x + x < 720

3x < 720

x < 720/3

x < 240

Since x should be less than 240,it means that the most that they can spend in food is $239

Answer

I think you meant to wtite

80.4/d = 0.0804

the answer is d = 1000 or 10^3

Step-by-step explanation: