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3 years ago

Which of the three equations have infinitely many solutions?

Mathematics

2 answers:

mina [271]3 years ago

6 0

Answer:

I need to know the questions luv.

Step-by-step explanation:

mr Goodwill [35]3 years ago

3 0

Answer:

Solving the three equations in part A showed that two equations have one solution and one equation has no solution. So, no equations have infinitely many solutions.

Step-by-step explanation:

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Bus stop Restaurant sell 3/4 as many cheeseburgers as they do with hamburgers on any given day. If there are a total of 42 burge

schepotkina [342]

Answer:

24cb 18 hb

Step-by-step explanation:

Let's call C the number of cheese burger sold and H the number of hamburgers.

As the cheeseburger sold are 3/4 of hamburgers, we can represent with the following equation:

C= (3/4)H (equation 1)

The total of burgers sold are the total of cheeseburger plus the total of hamburgers, and the sum 42. The equation representing it is:

C + H = 42

Replacing C by its value of eq. 1:

(3/4)H + H = 42

(3/4)H + (4/4)H = 42

(7/4) H = 42

Dividing both sides by (7/4)

H = 42/(7/4)

H= 42*(4/7)

H= (42/7)*4

H= 6*4

H=24

Then, C is:

C=(3/4)H

C=(3/4)*24

C= 3*24/4

C=72/4

C=18

We can verify it summing C+H

C+H=24+18=42 verified!

6 0

3 years ago

Expand 4x(7x-11) For maths Help please

harina [27]

Answer:

$28x^2-44x$

Step-by-step explanation:

Use the distributive property.

$4x * 7x = 28x^2$

$4x * (-11) = -44x$

We will get $28x^2-44x$

Hope this helps!

7 0

3 years ago

PLEASe PLEASE HeLP PLEASEEEEE

MA_775_DIABLO [31]

Answer:

option A

x=15/2

Step-by-step explanation:

refer to the above attachment

sorry for the messy writing:^

3 0

3 years ago

You are carpeting a room in a house for your job. You have a boss that likes to play jokes, so instead of just giving you the di

forsale [732]

Answer:

$x^2-4x+3$

$63\ \text{sq. units}$

Step-by-step explanation:

l = Length = $\dfrac{x^2-x-6}{x-5}$

w = Width = $\dfrac{x^2-6x+5}{x+2}$

Area is given by

$A=lw\\\Rightarrow A=\dfrac{x^2-x-6}{x-5}\times\dfrac{x^2-6x+5}{x+2}\\\Rightarrow A=\dfrac{x^2-3x+2x-6}{x-5}\times\dfrac{x^2-5x-1x+5}{x+2}\\\Rightarrow A=\dfrac{(x-3)(x+2)}{x-5}\times\dfrac{(x-5)(x-1)}{x+2}\\\Rightarrow A=(x-3)(x-1)\\\Rightarrow A=x^2-4x+3$