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3 years ago

How would i graph y=− 5/2x-3 and x−y=−4 (_,_)

Mathematics

1 answer:

SashulF [63]3 years ago

6 0

Given:

The system of equations is

$y=-\dfrac{5}{2}x-3$

$x-y=-4$

To find:

The graph of the given system of equations and its solution.

Solution:

The table of values for $y=-\dfrac{5}{2}x-3$ is

x y

-4 7

-2 2

0 -3

Plot these points and connect them by a straight line.

The table of values for $x-y=-4$ is

x y

-4 0

-2 2

0 4

Plot these points and connect them by a straight line.

From the below graph it is clear that the lines intersect each other at point (-2,2).

Therefore, the solution of given system of equation is (-2,2).

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Genrish500 [490]

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Hope this helps

3 0

2 years ago

A trailer will be used to transport several 40-pound crates to a store. The greatest amount of weight that can be loaded into th

docker41 [41]

Given :

A trailer will be used to transport several 40-pound crates to a store.

The greatest amount of weight that can be loaded into the trailer is 1,050 pounds.

An 82-pound crate has already been loaded onto the trailer.

To Find :

The greatest number of 40-pound crates that can be loaded onto the trailer.

Solution :

Weight left = 1050 - 80 = 970 pound.

Let, number of 40 pounds crates that can be loaded are x.

$x = \dfrac{970}{40}\\\\x = 24.25$

Since, crate cannot be in fraction, so maximum crate that can be loaded is 24.

Hence, this is the required solution.

7 0

2 years ago

HELP!! Can someone help me with this

cricket20 [7]

Answer:

1 no

2yes

3 no

4 no

5 Yes

6 Yes.

so it gives you a pattern right there a squared plus b squared equals c squared just fill in the numbers in that way

6 0

2 years ago

PLEASE HELP 25 POINTS!!

eduard

Answer:

(27 y^(-8))/(4 x^6)

Step-by-step explanation:

Simplify the following:

(4 (3 x^2 y^4)^3)/(2 x^3 y^5)^4

Multiply each exponent in 2 x^3 y^5 by 4:

(4 (3 x^2 y^4)^3)/(2^4 x^(4×3) y^(4×5))

4×5 = 20:

(4 (3 x^2 y^4)^3)/(2^4 x^(4×3) y^20)

4×3 = 12:

(4 (3 x^2 y^4)^3)/(2^4 x^12 y^20)

2^4 = (2^2)^2:

(4 (3 x^2 y^4)^3)/((2^2)^2 x^12 y^20)

2^2 = 4:

(4 (3 x^2 y^4)^3)/(4^2 x^12 y^20)

4^2 = 16:

(4 (3 x^2 y^4)^3)/(16 x^12 y^20)

Multiply each exponent in 3 x^2 y^4 by 3:

(4×3^3 x^(3×2) y^(3×4))/(16 x^12 y^20)

3×4 = 12:

(4×3^3 x^(3×2) y^12)/(16 x^12 y^20)

3×2 = 6:

(4×3^3 x^6 y^12)/(16 x^12 y^20)

3^3 = 3×3^2:

(4×3×3^2 x^6 y^12)/(16 x^12 y^20)

3^2 = 9:

(4×3×9 x^6 y^12)/(16 x^12 y^20)

3×9 = 27:

(4×27 x^6 y^12)/(16 x^12 y^20)

4/16 = 4/(4×4) = 1/4:

(27 x^6 y^12)/(4 x^12 y^20)

Combine powers. (27 x^6 y^12)/(4 x^12 y^20) = (27 x^(6 - 12) y^(12 - 20))/4:

(27 x^(6 - 12) y^(12 - 20))/4

6 - 12 = -6:

(27 x^(-6) y^(12 - 20))/4

12 - 20 = -8:

Answer: (27 y^(-8))/(4 x^6)

5 0

3 years ago

Two ropes, AD and BD, are tied to a peg on the ground at point D. The other ends of the ropes are tied to points A and B on a fl

Lisa [10]

Answer: 5(√3-1) unit.

Step-by-step explanation:

Since, By the below diagram,

For triangle BDC,

We can write,

$tan 30^{\circ}=\frac{BC}{DC}$

⇒ $\frac{1}{\sqrt{3}}=\frac{BC}{5\sqrt{3}}}$

⇒ $\sqrt{3}=\frac{5\sqrt{3}}{BC}$

⇒ $BC=\frac{5\sqrt{3} }{\sqrt{3} }$

⇒ $BC=5\text{ unit}$

Now, In triangle ADC,

$tan45^{\circ}=\frac{AC}{DC}$

⇒ $1=\frac{AB+BC}{DC}$

⇒ $1=\frac{DC}{AB+BC}$

⇒ $1=\frac{5\sqrt{3}}{AB+5}$

⇒ $AB+5=5\sqrt{3}$

⇒ $AB=5\sqrt{3}-5=5(\sqrt{3}-1)\text{ unit}$

8 0

3 years ago

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