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3 years ago

What equation represents the proportional relationship displayed in the table?

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

3 0

Answer:

$y=6x$

Step-by-step explanation:

Given:

The table is given as:

x 0 4 7 8

y 0 24 42 48

The relationship between 'x' and 'y' is a proportional relationship.

A proportional relationship is one in which one variable varies directly with the other by a factor of 'k' also known as constant of proportionality and $k\ne 0$

A proportional relationship is given as:

$y=kx$

Now, we plug in 'x' and 'y' values and find the value of 'k'.

Let $x=4\ and\ y=24$

$24=k\times 4\\\\k=\frac{24}{4}=6$

Therefore, the constant of proportionality is 6. Hence the complete equation is given as:

$y=6x$

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Find the value of x.

slavikrds [6]

Answer:

$x=11$

Step-by-step explanation:

we know that

The <u><em>Midpoint Theorem</em></u> states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

In this problem

$(x+5)=\frac{1}{2}(3x-1)$

solve for x

Multiply by 2 both sides

$2x+10=3x-1$

subtract 2x both sides

$10=3x-1-2x$

$10=x-1$

Adds 1 both sides

$x=11$

3 0

2 years ago

In the △ABC, the height AN = 24 in, BN = 18 in, AC = 40 in. Find AB and BC. Consider all possible cases. Case 1 : N∈BC, AB = __,

aivan3 [116]

Answer:

Case 1:

$AB = 30$

$BC = 50$

Case 2:

$AB = 15.9$

$BC = 36.7$

Case 3: Not possible

Step-by-step explanation:

Given

See attachment for illustration of each case

Required

Find AB and BC

Case 1:

Using Pythagoras theorem in ANB, we have:

$AB^2 = AN^2 + BN^2$

This gives:

$AB^2 = 24^2 + 18^2$

$AB^2 = 576 + 324$

$AB^2 = 900$

Take square roots of both sides

$AB = \sqrt{900$

$AB = 30$

To calculate BC, we consider ANC, where:

$AC^2 = AN^2 + NC^2$

$40^2 = 24^2 + NC^2$

$1600 = 576 + NC^2$

Collect like terms

$NC^2 = 1600 - 576$

$NC^2 = 1024$

Take square roots

$NC = \sqrt{1024$

$NC = 32$

So:

$BC = NC + BN$

$BC = 32 + 18$

$BC = 50$

Case 2:

Using Pythagoras theorem in ANB, we have:

$AN^2 = AB^2 + BN^2$

This gives:

$24^2 = AB^2 + 18^2$

$576 = AB^2 + 324$

Collect like terms

$AB^2 = 576 - 324$

$AB^2 = 252$

Take square roots of both sides

$AB = \sqrt{252$

$AB = 15.9$

To calculate BC, we consider ABC, where:

$AC^2 = AB^2 + BC^2$

$40^2 = 252 + BC^2$

$1600 = 252 + BC^2$

Collect like terms

$BC^2 = 1600 - 252$

$BC^2 = 1348$

Take square roots

$BC = \sqrt{1348$

$BC = 36.7$

Case 3:

This is not possible because in ANC

The hypotenuse AN (24) is less than AC (40)

3 0

2 years ago

Solve <br> 1. x2 + x - 6 = 0

ololo11 [35]

Answer:

x = -3 x=2

Step-by-step explanation:

x^2 + x - 6 = 0

Factor

What two numbers multiply to -6 and add to 1

3*-2 =-6

3+-2 =1

(x+3) (x-2) = 0

Using the zero product property

x+3 = 0 x-2 = 0

x = -3 x=2

6 0

2 years ago

PB_WITH_FSA_DIAG_6825_OL_FY21

astra-53 [7]

Answer:

uhm.....what?

Step-by-step explanation:

8 0

2 years ago

BRAINLIESTTT ASAP! PLEASE HELP ME :)

11Alexandr11 [23.1K]

<h3>Answer: B) Only the first equation is an identity</h3>

========================

I'm using x in place of theta. For each equation, I'm only altering the left hand side.

Part 1

cos(270+x) = sin(x)

cos(270)cos(x) - sin(270)sin(x) = sin(x)

0*cos(x) - (-1)*sin(x) = sin(x)

0 + sin(x) = sin(x)

sin(x) = sin(x) ... equation is true

Identity is confirmed

---------------------------------

Part 2

sin(270+x) = -sin(x)

sin(270)cos(x) + cos(270)sin(x) = -sin(x)

-1*cos(x) + 0*sin(x) = -sin(x)

-cos(x) = -sin(x)

We don't have an identity. If the right hand side was -cos(x), instead of -sin(x), then we would have an identity.

7 0

2 years ago