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2 years ago

Which equation(s) represents a proportional relationship? A) y = 5+x B) y = 4x C) y = 7x? D) y = 2-x

Mathematics

1 answer:

andreyandreev [35.5K]2 years ago

6 0

Answer:

B & C

Step-by-step explanation:

A proportional relationship has a proportionality constant (constant rate of change) and and must cross through the origin (or begin at (0,0). This means it's y-intercept is 0. Each of the options is a linear function. All of the functions have a constant rate of change. Only one as b=0 or 0 as the y-intercept.

B) y=4x has a constant of 4 and passes through the origin.

C) y=7x has a constant of 7 and passes through the origin.

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please answer please!!!! Will mark brainliest for complete answer. -the first photo is to calculate the area. Just question numb

cricket20 [7]

Triangle:
Area = 1/2*b*h
= (1/2)*3*4
= 3*2
= 6 m^2

7a.
1cm^3
Or, 1cm * 1cm * 1cm
Or, 10mm * 10 mm * 10 mm
Or, 1000 mm^3

7b.i)
10 cm^3
Or, 10cm * 1cm* 1cm
Or, 100mm * 10mm * 10mm
Or, 10000 mm^3

8.)
Volume = 50 cm^3
Base area= 5 cm^2
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5 0

2 years ago

Can some help me fast!

Mice21 [21]

Answer:

5x +2y =38

5(6) +2y = 38

30 +2y= 38

2y= 8

y= 4

(6, 4)

5(0) +2y =38

2y= 38

y= 19

(0,19)

5(-2) +2y= 38

-10 +2y =38

2y= 48

y=24

(-2,24)

7 0

2 years ago

Does 5n = -20 have one solution,infinitley many,or no solutions?

Maslowich

One solution.

-20 divided by 5 is equal to -4. Unless both sides have the same number or one side has -=0 then it’s one solution

8 0

2 years ago

Rewrite the system of linear equations as a matrix equation AX = B.

iren2701 [21]

Answer:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]$

And the vector of vriables (X) is:

$\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]$

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

$\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

In conclusion, AX=B is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

7 0

3 years ago

Simplify The Expression:<br><br> 1/8h+7-3/4h<br><br> Thank you!

ivann1987 [24]

Answer:

-5/8h + 7

Step-by-step explanation:

First you need to combine like terms (1/8h and -3/4h)

So 1/8h - 3/4h is -5/8h

Then we have our 7 left over

So the answer is -5/8h + 7

4 0

2 years ago