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3 years ago

Check the following pair of linear equations is consistent or inconsistent

Mathematics

2 answers:

alexandr1967 [171]3 years ago

8 0

Answer:

Consistent

Step-by-step explanation:

It was a solution.

Naya [18.7K]3 years ago

8 0

5x-3y=11
So first you need to switch -3y with 11 first because the the formula is y=mx+c
So, 5x -11= 3y
Now make y as the subject: 3y= 5x-11
Now divide both of the numbers by 3 to make it y=mx+c
So, y=3/5x-11
y=1.67x-3.67

Now try the another one by yourself by using my steps and remember that when you are switching the side. suppose if 3=5x if I will switch the side for this it will move from positive to negative so it will be -5x=-3.
If you want to switch from negative to positive it will be 3=5x (this is just an example)

I hope you understood my explanation. Thank me later.

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What is the volume of this sphere? use 3.14 for pi and round to the nearest hundredth when necessary.

Harrizon [31]

The volume of the sphere is 900 cm³

Step-by-step explanation:

Step 1: Given the diameter of the ball = 12 cm. Find radius.

Radius = diameter/2 = 12/2 = 6cm

Use the formula for volume of a sphere to find the volume.

Volume of the ball = 4/3 πr³

= 4/3 × 3.14 × (6)³

= 904.32 cm³ ≈ 900 cm³ (nearest hundredth)

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4 years ago

The luxury Swiss Chalet hotel general manager (GM) reported to her owner that the hotel's Occupancy Index for the calendar year

SpyIntel [72]

Answer:

the Swiss Chalet had higher occupancy than its competitive set in 2019

Step-by-step explanation:

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3 years ago

A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t2 + 6

galina1969 [7]

<h2>Hello!</h2>

The answers are:

a) The height of the ball after 3 seconds is 49 feet.

b) The maximum height of the ball is 66 feet.

c) That the ball hit the ground after 4 seconds.

d) The domain would be only the positive real numbers, from 0 to 4, since we found that the ball hit the ground at t equal to 4 seconds. However, if we were talking about a quadratic function with no time involved, the domain would be all the real numbers.

Since we are working with a quadratic function which describes the ball's motion in function of the time, we need to remember the following:

- The general equation of the parabola is:

$y=ax^{2} +bx+c$

- We can calculate the coordinates of the vertex of the parabola using the following formula:

$x_{vertex}=\frac{-b}{2a}$

- Evaluating a function means replacing the variable with the given value to evaluate.

The given function is:

$h(t)=-16x^{2}+63t+4$

Where,

$a=-16\\b=63\\c=4$

Now, calculating we have:

a) What is the height of the ball after 3 seconds?

We need to evaluate the time of 3 seconds into the function, so:

$h(3)=-16(3)^{2} +63(3)+4=49feet$

So, the height of the ball after 3 seconds is 49 feet.

b) What is the maximum height of the ball?

Since the function is describing the motion of a ball thrown into the air, we can find the maximum height by finding the y-coordinate of the vertex. If the parabola opens downward or upward, the vertex will be always the highest or the lowest point of the parabola.

So, calculating the vertex we have:

$x_{vertex}=\frac{-b}{2a}=\frac{-63}{2*-16}\\\\x_{vertex}=\frac{-63}{2*-16}=\frac{-63}{-32}=1.97$

Then, replacing "x" into the equation of the parabola, we find the y-coordinate of the vertex:

$y=-16(1.97)^{2}+63(1.97)+4=-16*3.88+63*1.97+4\\y=-16*3.88+63*1.97+4=-62.08+124.11+4=66.03$

So, if the y-coordinate is 66.03, the maximum height of the ball is 66.03 feet, or 66 feet (rounded to the nearest foot).

c) When will the ball hit the ground?

We can find the time when the ball hit the ground by making equal to 0 the function and finding the roots (zeroes)

Since it's a quadratic function, we can find the zeroes using the quadratic equation:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}$

Substituting a, b and c, we have:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\frac{-63+-\sqrt{63^{2}-4*(-16)*(4)} }{2*(-16)}\\\\\frac{-63+-\sqrt{63^{2}-4*(-16)*(4)} }{2*(-16)}=\frac{-63+-\sqrt{3969+256} }{-32}\\\\\frac{-63+-\sqrt{3969+256} }{-32}=\frac{-63+-\sqrt{4225} }{-32}=\frac{-63+-(65)}{-32}\\\\t1=\frac{-63-(65)}{-32}=4\\\\t1=\frac{-63+(65)}{-32}=-0.06$

Now, since negative time does not exists, we can conclude that the ball hit the ground after 4 seconds.

d) what domain makes sense for the function?

Since the function represents the motion of the thrown ball at "t" time, the domain would be only the positive real numbers, from 0 to 4, since we found that the ball hit the ground at t equal to 4 seconds. However, if we were talking about a quadratic function with no time involved, the domain would be all the real numbers,

Note: I've attached the graph of the function.

Have a nice day!

5 0

3 years ago

21 Maria sold small boxes of candy for $1 and

Anna35 [415]

Answer:

There where 10 <u>more</u> Small boxes sold than Large boxes.

Step-by-step explanation:

<u>This is a typical question that can be solved by obtaining a system of two linear equations and solving for each variable. </u>

In principle Linear Equations are algebraic expressions denoting a relationship between a Dependent variable $y$ and an Independent variable $x$ . In a system of Two Linear equations we have<u> two equations</u> of the same variable sets (thus Two Independent variables) so in this case both $y$ and $x$ will be variable terms.

Now with respect to the question and the given information, here our two Variable terms will be the small and the large boxes.

<u>Given Information:</u>

Small Boxes (lets call them $s$ ) cost $1 per box

Large Boxes (lets call them $l$ ) cost $4 per box
Total Number of Boxes sold is 30
Total Profit from sold Boxes is $60

Thus from the above we can obtain one equation denoting the Total Number of Boxes sold and one equation denoting the Total Profit from sold boxes, respectively, as follow:

$s+l=30$ Eqn(1): Total Number of Boxes

$1s+4l=60$ Eqn(2): Total Profit from sold boxes

Now we have a system of two linear equations which we can solve and find the number of small and large boxes, $s$ and $l$ respectively.

From Eqn(1) we see that

$s=30-l$ Eqn(3).

Plugging Eqn(3) in Eqn(2) we can solve for $l$ as:

$1(30-l)+4l=60$

$30-l+4l=60$ Factored out bracket

$3l=60-30$ Gather similar terms on each side and simplify

$3l=30$

$l=\frac{30}{3}$

$l=10$

Plugging in the value for $l=10$ in Eqn(3) we have

$s=30-10\\s=20$

So we know that they were<u> 10 Large Boxes and 20 Small Boxes sold</u>, thus to answer our question, there where 10 more Small boxes sold than Large boxes.

5 0

4 years ago

4 chairs and 3 tables together cost ₨ . 2100 and 5 chairs and 2 tables cost Rs.1750. Find the cost of 1 chair and 1 table while

malfutka [58]

Answer:

Cost of one chair = Rs . 150

Cost of one table = Rs. 500

Step-by-step explanation:

Let number of chairs denote x and no of tables denote y

4x + 3y = 2100 --------------------(i)

5x +2y = 1750 ------------------(ii)

Multiply equation (i) by 2 and (ii) by (-3) and now y will be eliminated.

(i)*2 8x + 6y = 4200

(ii)*(-3) -<u>15x - 6y = - 5250</u> {Now, add}

- 7x = - 1050

x = -1050/-7

x = Rs. 150

Plug in the value of x inn equation (i)

4*150 + 3y = 2100

600 + 3y = 2100

3y = 2100 - 600

3y = 1500

y= 1500/3

y =Rs 500

5 0

3 years ago