Question 2 of 11
What system of equations would you use to solve the problem below?
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Using proportions, the estimate of the per capita consumption from 2011 is of 29.2 gallons.

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

From the data, the 2012 consumption was 5.3% percent higher than in 2011, that is 105.3% of 1.053x. This consumption was of 30.8 gallons, hence:

1.053x = 30.8.

x = 30.8/1.053

x = 29.2.

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Answer:

[tex]3 \frac{7}{8}[/tex]

Step-by-step explanation:

Answer:

x = 99

Step-by-step explanation:

3x + 99x = 12x

12x = 1188 divide 1188 ÷ 12

x = 99

Answer:

64%

Step-by-step explanation:

Convert 16/25 to Percentage by Converting to Decimal

With this method, we first need to divide the numerator by the denominator:

16 ÷ 25 = 0.64

Once we have the fraction in a decimal format, the answer is then multiplied by 100 to get the correct percentage:

0.64 × 100 = 64%

We can see that this gives us the exact same answer as the first method: 16/25 as a percentage is 64%.

Note, the final percentage is rounded to 2 decimal places to make the answer simple to read and understand.

Answer:

7x-1 + 2x-1 =7 ..... given

9x-2 = 7

9x = 9

x = 1

Step-by-step explanation:

This is the answer of the above condition.

Please mark as BRAINLIEST and THANK ME.

Answer:

y = 4x - 5

Step-by-step explanation:

y = mx + b

By putting above values in equation (i)

-1 = 4 (1) + b

-1 = 4 + b

-5 = b

b = -5

y = 4x - 5

Answer:

[tex]\sf 1. \quad \tan(B)=\dfrac{4}{3}[/tex]

[tex]\sf 2. \quad \tan(A)=\dfrac{7}{2}[/tex]

[tex]\sf 3. \quad \angle A[/tex]

Step-by-step explanation:

Acute angle: an angle less than 90°

Tan trigonometric ratio

[tex]\sf \tan(\theta)=\dfrac{O}{A}[/tex]

where:

Question 1

Refer to the first attached diagram.

If tan(A) = 3/4 then the side opposite ∠A is 3 units, and the side adjacent ∠A is 4 units.

Therefore:

[tex]\implies \sf \tan(B)=\dfrac{4}{3}[/tex]

Question 2

Refer to the second attached diagram.

If tan(B) = 2/7 then the side opposite ∠B is ,and the side adjacent ∠B is 7.

Therefore:

[tex]\implies \sf \tan(A)=\dfrac{7}{2}[/tex]

Angle A would be bigger (see second attachment).

Answer:

  (a)  y = 5/12x and y = -5/12x

Step-by-step explanation:

Given the equation of a hyperbola 25x² – 144y² + 3,600 = 0, you want the equations of the asymptotes.

Dividing by 3600 and doing a little rearrangement, we have ...

  (y/5)² -(x/12)² = 1 . . . . . standard form equation of a hyperbola

The equation of the asymptotes is ...

  (y/5)² -(x/12)² = 0

Solving for y gives ...

  (y/5)² = (x/12)²

  y/5 = ±x/12 . . . . . . take the square root

  y = ±5/12·x . . . . . . . matches the first choice

__

Additional comment

The equation of a hyperbola with vertices on the y-axis is often written as ...

  y²/a² -x²/b² = 1

We have written it in the above form to better show the scale factors applied to x and y.

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Answer:

Step-by-step explanation:

Find the area of a circle with a diameter of 10 ft.

Area of a Circle  

A = π × r²

r = diameter : 2

10 : 2 = 5 ft (radius)

A = π × 5²

A = 3.14 * 25

A = 78.5 ft²

Answer:

20x + 8

Step-by-step explanation:

5*4x + 5*5 -17

20x +25 -17

20x + 8

Answer:

16 x^14 / y^16

Step-by-step explanation:

Remember   ' PEDMAS '

Parentheses

Exponents

Division

Multiplication

Addition

Subtraction

 so do the stuff inside the parentheses to get

  y^8     ^-2

  4x^7  

Now do the exponents  ( the -2  one)

y^-16

4^-2 x^-14       then simplify

16 x^14 / y^16

The solution is  x ≈ 0.3 , Option A is the right Answer.

An Equation is a mathematical statement formed when two algebraic expressions are equated using an equal sign.

To find the value of

22x = 5.3

A graph has to be plotted for the line

From the graph it can be clearly seen the point at which the line intersect x axis is  x = 0.24 , which is approx equal to 0.3.

Therefore the solution is  x ≈ 0.3 ,

Option A is the right Answer.

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Answer:

3 cakes

Step-by-step explanation:

just flow by the formula then ignore the remainder and take the whole number which is 3 then that becomes your answer good day.

The no. of different groups of players of 6 players , if the position does not matter is given by 5005 ,Option C is the right answer.

Combination is a method of selection of items from a set of distinct objects , when the position doesn't matter .

It is given that

Total number of players = 15 = n

No. of players to be selected = 6 = r

The no. of different groups of players of 6 players , if the position does not matter is given by

[tex]\rm ^nC_r = \dfrac{n!}{r! (n-r)!}[/tex]

So

[tex]\rm ^{15}C_6 = \dfrac{15!}{6! (15-6)!}\\\\\rm ^{15}C_6 = \dfrac{15!}{6! (9)!}\\\\\\^{15}C_6 = \dfrac{15 * 14 * 13* 12 * 11 * 10 }{6 *5* 4 * 3 * 2*1}\\\\ ^{15}C_6 = 5005\\[/tex]

Therefore , The no. of different groups of players of 6 players , if the position does not matter is given by 5005

Option C is the right answer.

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Step-by-step explanation:

1/2 represents every fraction, where the denominator (bottom part) is twice the numerator (top part).

like 4/8 or 6/12 or 128/256 or ...

what do we need in the denominator to calculate with 15/16 ?

the same : 16.

and what is half of 16 ? 8.

so, we need the 1/2 based on 16th = 8/16.

so, the dog ate 8/16 (1/2) of the original 15/16.

what was left was

15/16 - 8/16 = 7/16 pound

Answer:

4 kg

Step-by-step explanation:

so 1.6 * 4 is 6.4 so you just double check if you want

4 + 1.6 which is 1/4 of 6.4 will be 5.6 (matches with the question)

4 + (1.6*4) (indicates it's full) = 10.4( also Matches with the question)

there is another way by setting up an equation. let me know in comments if you want to see it that way

If  tank weighs 5.6kg when it 1/4 filled with water. If it weighs 10.4kg when it is full , then weight when it is empty is 0.93kg

Two or more expressions with an Equal sign is called as Equation.

Given,

A tank weighs 5.6kg when it 1/4 filled with water

Tank weighs 10.4kg when it is full

We need to find the weight when it is empty.​

Let weight of the tank “x” and the weight of the water in the full tank “y”.

From the problem statements, we deduce the following equations:

x + y/4 = 5.6kg => x = 5.6kg - y/4

x + y = 10.4kg => y = 10.5kg - x

y = 10.5kg - (5.6kg - y/4)

3y/4 = 4.9

Thus y = 6.53kg; x = 0.93kg

Hence, weight of the tank when it is empty is 0.93kg.

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It can be modeled by an exponential function because each time x increases by 1, f(x) is multiplied by 4.

This means the base of the exponential function is 4, and since f(x)=3 when x=0, [tex]f(x)=3(4)^{x}[/tex]

Answer: 1, -2, 1

Step-by-step explanation:

[tex]f(g(x))=f(x-1)=(x-1)^{2}=x^{2}-2x+1[/tex]

To find the point of intersection for two lines, we can use substitution to solve for the coordinates.

We're given:

With these two equations, we can:

Isolate x in the second equation:

[tex]x+2y=7\\x =7-2y[/tex]

Substitute the second equation into the first:

[tex]3x-2y=5\\3(7-2y)-2y=5\\21-6y-2y=5\\21-8y=5\\21-8y=5[/tex]

Solve for y:

[tex]21-6y-2y=5\\21-8y=5\\-8y=-16\\y=2[/tex]

Solve for x:

[tex]x+2y=7\\x+2(2)=7\\x+4=7\\x=3[/tex]

The point of intersection of the lines is (3,2).

Answer:

False

Step-by-step explanation:

P is not necessarily on the angle bisector of angle A.

Attached is an example of a carefully constructed shape (so that it's easy to see the measurements):

Visually, point P is not on the angle bisector of angle A.

While having two adjacent congruent segments is necessary for P to be on the bisector of angle A, it is not sufficient.  The segments on the rays from A to the end of those segments would also need to be a pair of congruent adjacent sides (forming a kite).  For a kite, the diagonal from the point between one pair of adjacent congruent sides to the point between the other pair of adjacent congruent sides does form an angle bisector with both angles.

The pressure by the statue exerted on the floor will be 1200 N per square meter.

The term pressure is defined as the force per unit area.

The formula is igven below.

P = F /A

A bronze statue weighs 2400 N and has a base that is 4 m by ½ m.

Then the base area will be

A = 4 x 1/2

A = 2 square meters

Then the pressure will be

P = 2400 / 2

P = 1200 N per square meter.

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Answer:

C

Step-by-step explanation:

A and B are wrong because 37 is prime. 31 is prime and 33 is divisible by 11 and 3 so C

Answer:

C

Step-by-step explanation:

The slope is -1/3, so it will go down and to the right. Down 1, to the right 3 (rise over run!)

The y intercept is 2.

Answer:

The answer is either 30 degrees or 36 degrees.
You need a protractor to measure it.
Most probably, the answer is 30 degrees.
BUT PLEASE MEASURE THE ANGLE WITH A PROTRACTOR!

Answer:

d. 60°

Step-by-step explanation:

Since the clock is a full circle, it has a degree span of 360°.

∴ every 5-min interval on the clock

= 360° ÷ 12 five min intervals

= 30°

∴ acute angle formed by clock in figure

= two 5-min intervals

= 60°

Based on the altitudes of both the airplane and the helicopter, the number of minutes until they are at the same altitude is 20 minutes.

Assuming the minute both will be on the same altitude is x, the altitude that the airplane would be at that point is:

= 30,000 - 1,000x

The helicopter's altitude would be:

= 0 + 500x

Putting them together gives:

30,000 - 1,000x = 500x

30,000 = 1,500x

x = 30,000 / 1,500

= 20 minutes

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Answer:

C.) y = 3(x - 3)(x - 7)

For the function:

Writing equation:

y = a(x - h)² + k

15 = a(8 - 5)² - 12

15 = 9a - 12

9a = 15 + 12

9a = 27

a = 3

So the equation is:

y = 3(x - 5)² - 12

y = 3(x² - 10x + 25) - 12

y = 3x² - 30x + 63

y = 3(x - 3)(x - 7)

Answer:

4/7

Step-by-step explanation:

2/3 × 6

7

4/7

and bro that's really it

(a) If [tex]\alpha[/tex] and [tex]\beta[/tex] are roots of [tex]f(x)[/tex], then we can factorize [tex]f[/tex] as

[tex]f(x) = x^2 + (k - 6) x + 9 = (x - \alpha) (x - \beta)[/tex]

Expand the right side and match up coefficients:

[tex]x^2 + (k-6) x + 9 = x^2 - (\alpha + \beta) x + \alpha \beta \implies \begin{cases} \alpha + \beta = -(k-6) \\ \alpha \beta = 9 \end{cases}[/tex]

Now, recall that [tex](x+y)^2 = x^2 + 2xy + y^2[/tex]. It follows that

[tex]\boxed{\alpha^2 + \beta^2} = (\alpha + \beta)^2 - 2\alpha\beta = (-(k-6))^2 - 2\times9 = \boxed{k^2 - 12k + 18}[/tex]

and

[tex]\boxed{\alpha^2\beta^2} = 9^2 = \boxed{81}[/tex]

(b) If [tex]9(\alpha^2+\beta^2) = 2\alpha^2\beta^2[/tex], then

[tex]9 (k^2 - 12k + 18) = 2\times81 \implies 9k^2 - 108k = 0 \implies 9k (k - 12) = 0[/tex]

Since [tex]k\neq0[/tex], it follows that [tex]\boxed{k=12}[/tex].

(c) The simplest quadratic expression with roots [tex]\frac1{\alpha^2}[/tex] and [tex]\frac1{\beta^2}[/tex] is

[tex]\left(x - \dfrac1{\alpha^2}\right) \left(x - \dfrac1{\beta^2}\right)[/tex]

which expands to

[tex]x^2 - \left(\dfrac1{\alpha^2} + \dfrac1{\beta^2}\right) x + \dfrac1{\alpha^2\beta^2}[/tex]

Reusing the identity from (a-i) and the result from part (b), we have

[tex]\left(\dfrac1\alpha + \dfrac1\beta\right)^2 = \dfrac1{\alpha^2} + \dfrac2{\alpha\beta} + \dfrac1{\beta^2} \\\\ \implies \dfrac1{\alpha^2} + \dfrac1{\beta^2} = \left(\dfrac{\alpha + \beta}{\alpha\beta}\right)^2 - \dfrac2{\alpha\beta} = \left(\dfrac{-(k-6)}9\right)^2 - \dfrac29 = \dfrac29[/tex]

We also know from part (a-ii) that [tex]\alpha^2\beta^2=81[/tex].

So, the simplest quadratic that fits the description is

[tex]x^2 - \dfrac29 x + \dfrac1{81}[/tex]

To get one with integer coefficients, we multiply the whole expression by 81 to get [tex]\boxed{81x^2 - 18x + 1}[/tex].