MATHEMATICS HIGH SCHOOL

Select all statements that are true about the linear equation.y=4x−3

The graph of the equation is a single point, representing one solution to the equation.

The point (1, 1) is on the graph of the equation.

4x−y=−3 has the same graph.

Since the point (0, −3) is a solution to the equation, it is on the graph of the equation.

The graph of the equation is the set of all points that are solutions to the equation.

Answers

Answer 1

Answer:

The correct answer for this question would be:

"The graph of the equation is the set of all points that are solutions to the equation."

"The point (1, 1) is on the graph of the equation."

And "The point (0, -3) is on the graph of the equation."

- I just took the test.

Answer 2

Answer: For point (1, 1): 1 = 4(1) - 3 = 4 - 3 = 1.
Therefore, the point (1, 1) is on the graph of the equation.

Since the point (0, −3) is a solution to the equation, it is on the graph of the equation.

The graph of the equation is the set of all points that are solutions to the equation.

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The sides of a quadrilateral are 3,4,5 and 6. Find the length of the shortest side of a similar quadrilateral whose area is 9 times as great. A) 9

B) 13.5

C) 27

Answers

Answer:

(A) $9$

Step-by-step explanation:

GIVEN: The sides of a quadrilateral are $3,4,5$ and $6$ .

TO FIND: Find the length of the shortest side of a similar quadrilateral whose area is $9$ times as great.

SOLUTION:

let the height of smaller quadrilateral be $h$

As both quadrilateral are similar,

let the length of larger quadrilateral are $x$ times of smaller.

sides of large quadrilateral are $3x,4x,5x\text{ and }6x$

height of large quadrilateral $=h x$

Area of lager quadrilateral $=\text{base}*\text{height}$

$=4x* hx=4hx^2$

Area of smaller quadrilateral $=\text{base}*\text{height}$

$=4h$

as the larger quadrilateral is $9$ times as great

$(4hx^2)/(4h)=9$

$x^2=9$

$x=3$

shortest side $=3x=3*3=9$

Hence the shortest side of larger quadrilateral is $9$ , option (A) is correct.

Answer:

(A)

Step-by-step explanation:

GIVEN: The sides of a quadrilateral are and .

TO FIND: Find the length of the shortest side of a similar quadrilateral whose area is times as great.

SOLUTION:

let the height of smaller quadrilateral be

As both quadrilateral are similar,

let the length of larger quadrilateral are times of smaller.

sides of large quadrilateral are

height of large quadrilateral

Area of lager quadrilateral

Area of smaller quadrilateral

as the larger quadrilateral is times as great

shortest side

Hence the shortest side of larger quadrilateral is , option (A) is correct.

Step-by-step explanation:

Help me..... please.....

Answers

Hello!
The first step is to add the 40 to both sides of the equation because the 40 is negative, and adding a 40 will make it 0.
Hope this helps!

Please help me asapp math is shi

Answers

Answer:

y= 15x+30

Step-by-step explanation:

Find slope using 2 points. Y intercept is where the graph intersects the y axis and you can see it intersects at 30

Y=15x + 30 :) hope this helps and all best of luck for your math

What are the solutions to the quadratic equation 4x2 = 64? A. x = −16 and x = 16 B.x = −8 and x = 8 C.x = −4 and x = 4 D.x = −2 and x = 2

Answers

Answer:

x = ±4

Step-by-step explanation:

4x^2 = 64

Divide each by 4

4x^2 /4= 64/4

x^2 = 16

Take the square root of each side

sqrt(x^2) = ±sqrt(16)

x = ±4

Answer:

$\boxed{\boxed{x=\pm 4}}$

Step-by-step explanation:

$4x^2 = 64$

Divide both sides by 4.

$(4x^2)/4 = 64/4$

Simplify.

$x^2 =16$

Take the square root on both sides.

$√(x^2 ) =\pm √(16)$

Simplify.

$x=\pm 4$

Y=2x+1 find the slope and y-intercept

Answers

$The\ slope-intercept\ form:y=mx+b\n\nm\ is\ the\ slope\nb\ is\ y-intercept\n=========================\ny=2x+1\n\nthe\ slope:m=2\n\ny-intercept:b=1$

y = mx + b

y=2x+1

Slope = 2
Y-Intercept = 1

What are the solutions to the system of equations graphed below?

Answers

Answer:

(0,-4) and (2,0)

Step-by-step explanation:

you cut off the screen shot, so not all answer options are visible.

but from the graph we can pretty reliably deduct, that the solutions (the crossing points of both functions) are

(0,-4) and (2,0)

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