Eliminate the fractions before solving for x. Explain each of your steps. 1/3+1/9x + 4/18x = 5/9

Answer:

24

Step-by-step explanation:

The numerators are being multiplied by -2

Answer:

yes

Step-by-step explanation:

If 2 lines are perpendicular then the product of their slopes = - 1

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - x - 4 ← is in slope- intercept form

with slope m = - 1

5x - 5y = 20 ( subtract 5x from both sides )

- 5y = - 5x + 20 ( divide all terms by - 5 )

y = x - 4 ← in slope- intercept form

with slope m = 1

Thus product of their slopes is - 1 × 1 = - 1

Therefore the lines are perpendicular

Answer:

there is no specific answer

Step-by-step explanation:

are we told to solve for n , simplify or factories

Answer:

64.5668 inches tall

Step-by-step explanation:

If you set up a proportion, you would have to do 1.64 times 39.37, which would equal 64.5668 inches.

Answer:

I'd say 3,tbh

Hope this helps :)

Answer:

x = 111°

Step-by-step explanation:

Angles in a triangle add up to 180

180 - (29+82) = 69

Angles on a straight line add up to 180

180 - 69 = 111

Answer:

111°

Step-by-step explanation:

the sum of the outer angles of a triangle is alway 180

so 180-82-29= 69°

180-69= 111°

Answer:

1st side= 4m

2nd side= 8m

3rd side= 12m

Step-by-step explanation:

Let the first, second and third side of the triangle be a, b and c meters respectively.

a= b -4 -----(1)

c= 3a -----(2)

Perimeter of triangle

= a +b +c

= 24

a +b +c= 24 -----(3)

From (1): b= a +4 -----(4)

Subst. (2) and (4) into (3):

a +(a +4) +3a= 24

3a +a +a +4= 24

5a +4= 24

5a= 24 -4

5a= 20

Divide both sides by 5:

a= 20 ÷5

a= 4

Substitute a= 4 into (4):

b= 4 +4

b= 8

Substitute a= 4 into (2):

c= 3(4)

c= 12

Thus, the lengths of the first, second, and third side of the triangle are 4m, 8m and 12m respectively.

The equation that relates the graph is a) y= - 1/2(x-3)²+2

Here Let put x=7

y= -1/2(x-3)²+2

= -1/2(7-3)²+2

= -1/2(4)²+2

= -1/2(16)+2

= - 8+2

= - 4

Hence (7,-4) is marked.

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The value that would complete the table to make the relationship between the two quantity proportional is 4.

When two quantities are proportional, their relationship is constant for all values and as one quantity rises, the other rises as well.

A proportional relationship exists between two quantities if they can be written in the general form y = kx, where k is the proportionality constant. In other words, the ratio between these amounts never changes. In other words, no matter which pair of the two numbers you divide, you always obtain the same number k.

The value that would complete the table to make the relationship between the two quantity proportional is 4.

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The system of inequalities in the graph is the one in option A.

y ≥ (-2/5)x - 2/5

y ≥ (3/2)*x - 1

Here we can see the graph of a system of inequalities, on the graph we can see two lines.

The first one is a line with a positive slope, it has an y-intercept of -1, the shaded region is above that line,  and it is a solid line, so one of the inequalities is:

y ≥ a*x - 1

Where a is positive.

The second line has a negative slope, and we can see that the shaded region is also above the line, so this second inequality is like:

y ≥ line with negative slope.

It is easy to identify the correct option because there is only one with these properties, which is the first option:

y ≥ (-2/5)x - 2/5

y ≥ (3/2)*x - 1

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Answer: 194

Step-by-step explanation:

From the volumes, we deduce that the side lengths of the cubes are 1 cm, 2 cm, 3 cm, and 5 cm. We position the cubes as follows:

[asy]

unitsize(0.5 cm);

draw((0,0)--(5*dir(-30))--(5*dir(-30) + 5*dir(30))--(10*dir(-30))--(5*dir(-30) + 5*dir(-90))--(5*dir(-90))--(0,0));

draw((5*dir(-30))--(5*dir(-30) + 5*dir(-90)));

draw((0,0)--(0,2)--((0,2) + 2*dir(-30))--(2*dir(-30)));

draw((0,2)--((0,2) + 2*dir(30))--((0,2) + 2*dir(30) + 2*dir(-30))--(2*dir(30)));

draw((2*dir(-30))--(2*dir(-30) + dir(30))--(2*dir(-30) + dir(30) + (0,1))--(2*dir(-30) + 2*dir(30) + (0,1))--(2*dir(-30) + 2*dir(30) + (0,2)));

draw((2*dir(-30) + dir(30))--(3*dir(-30) + dir(30))--(3*dir(-30) + dir(30) + (0,1))--(2*dir(-30) + dir(30) + (0,1)));

draw((3*dir(-30) + dir(30) + (0,1))--(3*dir(-30) + 2*dir(30) + (0,1))--(2*dir(-30) + 2*dir(30) + (0,1)));

draw((2*dir(30) + (0,2))--(2*dir(30) + (0,3))--(2*dir(30) + 3*dir(-30) + (0,3))--(2*dir(30) + 3*dir(-30))--(dir(30) + 3*dir(-30)));

draw((2*dir(30) + (0,3))--(5*dir(30) + (0,3))--(5*dir(30) + 3*dir(-30) + (0,3))--(5*dir(30) + 3*dir(-30))--(5*dir(30) + 5*dir(-30)));

draw((3*dir(-30) + 2*dir(30))--(3*dir(-30) + 5*dir(30)));

draw((3*dir(-30) + 2*dir(30) + (0,3))--(3*dir(-30) + 5*dir(30) + (0,3)));

[/asy]

The surface area of a cube with side length $s$ is $6s^2$, so the total surface area of the cubes is $6 \cdot 1^2 + 6 \cdot 2^2 + 6 \cdot 3^2 + 6 \cdot 5^2 = 234$.

Note that every pair of cubes touches, and furthermore, they have maximum contact. (This is why this solid has the smallest possible area.) The area of contact of the 1-cube and the 2-cube is 1 square centimeter, so we must subtract this twice from 234 (because this portion of the area from both the 1-cube and 2-cube is not seen anymore).

Doing this for every pair of cubes, we find that the surface area of this solid is $234 - 2 \cdot 1^2 - 2 \cdot 1^2 - 2 \cdot 1^2 - 2 \cdot 2^2 - 2 \cdot 2^2 - 2 \cdot 3^2 = \boxed{194}$.

Answer:

9x

Step-by-step explanation:

All we need to do is add like terms wich would be the 2s.

Next we would divided but sense -2+2 is 0 we just have x (which is 1) so our answer would be 9x

PLLLSS GIVE BRAINLIST MEAN THE WORLD

Question 1

Let \(x=18\) and \(y=25\). Then, we are required to compute:

\((x+y)^3 -x^3-y^3=x^3 +3x^2 y+3y^2 x+y^3-x^3-y^3=3xy(x+y)=\boxed{58050}\)

Question 2

Using the remainder theorem, the remainder is equal to \(x^4 -x^3 +2x+1\) computed at \(x=-1/2\). Doing so yields \((-1/2)^4 -(-1/2)^3 +2(-1/2)+1=\boxed{3/16}\)

Question 3

Using the factor theorem, the polynomial \(x^3 -2kx^2 +x+3\) should equal zero when evaluated at \(x=1\).

\(\implies 1-2k+1+3=0 \implies k=\boxed{5/2}\)

Answer:

3780

Step-by-step explanation:

10500 x 9% x 4 divided by 100=3780

\(\text{Plug in and solve:}\\\\3z\\\\3(9)\\\\\boxed{27}\)

Answer:

27

Step-by-step explanation:

3z       z=9

3 times 9

27

The ratio of the plant cell's length to its width is 2:1 (5.8 x 10^-6 m to 2.9 x 10^-6 m).

To calculate the ratio of the plant cell's length to its width, we first need to find the length and width of the cell. The given data states that the length of the cell is 5.8 x 10^-6 m and the width is 2.9 x 10^-6 m. To calculate the ratio, we divide the length by the width. In this example, we divide 5.8 x 10^-6 m by 2.9 x 10^-6 m, which equals 2. Thus, the ratio of the plant cell's length to its width is 2:1.

length / width = 5.8 x 10^-6 m / 2.9 x 10^-6 m = 2

length : width = 5.8 x 10^-6 m : 2.9 x 10^-6 m = 2:1

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Answer:

X=2, y=-1

Step-by-step explanation:

-4x +8y=-16 equation 1

4x +3y=5 equation 2

     11y=-11 add two equations to eliminate x

y=-1

solve for x by inputting y value into either equation

4x+3y=5

4x+3(-1)=5

4x-3=5

4x=8

x=2

Answer:

Quadrant III and IV

Step-by-step explanation:

Nonzero means any number but 0.

Given:

Steven wrote this equation:

\(4\times 18=418\)

To find:

The error in Steven's reasoning and then find the correct product.

Solution:

We have,

\(4\times 18=418\)

This statement is incorrect because we cannot write product directly as Steven wrote.

We need to multiply the place values of each digit of first number with the place values of second number and then we need to add the resulted values.

\(4\times 18=4\times (10+8)\)

\(4\times 18=4\times 10+4\times 8\)

\(4\times 18=40+32\)

\(4\times 18=72\)

Therefore, the correct product is 72.

a) To prove the proposition "For all sets S and T, SOTS," we need to show that for any sets S and T, S is a subset of the intersection of S and T.

To prove this, let's assume that S and T are arbitrary sets. We want to show that if x is an element of S, then x is also an element of the intersection of S and T.

By definition, the intersection of S and T, denoted as S ∩ T, is the set of all elements that are common to both S and T. In other words, an element x is in S ∩ T if and only if x is in both S and T.

Now, let's consider an arbitrary element x in S. Since x is in S, it is also in the set of all elements that are common to both S and T, which is the intersection of S and T. Therefore, we can conclude that if x is an element of S, then x is also an element of S ∩ T.

Since we've shown that every element in S is also in S ∩ T, we can say that S is a subset of S ∩ T. Thus, we have proved the proposition "For all sets S and T, SOTS."

b) To prove the proposition "For all sets S and T, S-TS," we need to show that for any sets S and T, S minus T is a subset of S.

To prove this, let's assume that S and T are arbitrary sets. We want to show that if x is an element of S minus T, then x is also an element of S.

By definition, S minus T, denoted as S - T, is the set of all elements that are in S but not in T. In other words, an element x is in S - T if and only if x is in S and x is not in T.

Now, let's consider an arbitrary element x in S - T. Since x is in S - T, it means that x is in S and x is not in T. Therefore, x is also an element of S.

Since we've shown that every element in S - T is also in S, we can say that S - T is a subset of S. Thus, we have proved the proposition "For all sets S and T, S-TS."

c) To prove the proposition "For all sets S, T, and W, (ST)-WES-(T- W)," we need to show that for any sets S, T, and W, the difference between the union of S and T and W is a subset of the difference between T and W.

To prove this, let's assume that S, T, and W are arbitrary sets. We want to show that if x is an element of (S ∪ T) - W, then x is also an element of T - W.

By definition, (S ∪ T) - W is the set of all elements that are in the union of S and T but not in W. In other words, an element x is in (S ∪ T) - W if and only if x is in either S or T (or both), but not in W.

On the other hand, T - W is the set of all elements that are in T but not in W. In other words, an element x is in T - W if and only if x is in T and x is not in W.

Now, let's consider an arbitrary element x in (S ∪ T) - W. Since x is in (S ∪ T) - W, it means that x is in either S or T (or both), but not in W. Therefore, x is also an element of T - W.

Since we've shown that every element in (S ∪ T) - W is also in T - W, we can say that (S ∪ T) - W is a subset of T - W. Thus, we have proved the proposition "For all sets S, T, and W, (ST)-WES-(T- W)."

d) To prove the proposition "For all sets S, T, and W, (T-W) nS = (TS)-(WNS)," we need to show that for any sets S, T, and W, the intersection of the difference between T and W and S is equal to the difference between the union of T and S and the union of W and the complement of S.

To prove this, let's assume that S, T, and W are arbitrary sets. We want to show that (T - W) ∩ S is equal to (T ∪ S) - (W ∪ S').

By definition, (T - W) ∩ S is the set of all elements that are in both the difference between T and W and S. In other words, an element x is in (T - W) ∩ S if and only if x is in both T - W and S.

On the other hand, (T ∪ S) - (W ∪ S') is the set of all elements that are in the union of T and S but not in the union of W and the complement of S. In other words, an element x is in (T ∪ S) - (W ∪ S') if and only if x is in either T or S (or both), but not in W or the complement of S.

Now, let's consider an arbitrary element x in (T - W) ∩ S. Since x is in (T - W) ∩ S, it means that x is in both T - W and S. Therefore, x is also an element of T ∪ S, but not in W or the complement of S.

Similarly, let's consider an arbitrary element y in (T ∪ S) - (W ∪ S'). Since y is in (T ∪ S) - (W ∪ S'), it means that y is in either T or S (or both), but not in W or the complement of S. Therefore, y is also an element of T - W and S.

Since we've shown that every element in (T - W) ∩ S is also in (T ∪ S) - (W ∪ S') and vice versa, we can conclude that (T - W) ∩ S is equal to (T ∪ S) - (W ∪ S'). Thus, we have proved the proposition "For all sets S, T, and W, (T-W) nS = (TS)-(WNS)."

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Therefore, the ladder touches the house at a height of 28 feet above the ground.

A triangle is a geometric shape that consists of three straight sides and three angles. It is a polygon with three sides. The sides of a triangle are connected by its vertices or corners. The triangle is one of the simplest and most fundamental shapes in geometry, and it has many important properties and applications. Triangles have many practical applications in everyday life and in various fields, such as architecture, engineering, and physics. The study of triangles and their properties is an important part of mathematics and geometry.

Here,

We can use the Pythagorean theorem to solve this problem. The Pythagorean theorem states that in a right triangle, the sum of the squares of the lengths of the legs (the two shorter sides) is equal to the square of the length of the hypotenuse (the longest side, which is opposite the right angle). In this problem, the ladder, the side of the house, and the ground form a right triangle. The ladder is the hypotenuse, the distance from the house to the ladder is one leg, and the height we want to find is the other leg.

Let x be the height above the ground where the ladder touches the house. Then, using the Pythagorean theorem, we have:

x² + 21² = 35²

Simplifying and solving for x, we get:

x² + 441 = 1225

x² = 784

x = 28

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Answer:

its cruked i dont see it

Step-by-step explanation:

Answer:

true

Step-by-step explanation: