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3 years ago

Which line has a constant of proportionality between y and x of 5/4

Mathematics

1 answer:

Paraphin [41]3 years ago

7 0

Answer:

Step-by-step explanation:

Because under the six there's a line so the line is there then you look at the four and equal it to it and that's how you know it's B

You're welcome

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Solve the equation .<br>a. 5×=35<br>b. 4× = 18<br>c. 6c–2c = 12.

gulaghasi [49]

Answer:

a. x=7 b. x= 4.5 c. c= 3

Step-by-step explanation:

a. 5x=35

5x/5=35/5

x = 7

b. 4x=18

4x/4=18/4

x=4.5

c. 6c-2c=12

4c = 12

4c/4 = 12/4

c=3

5 0

2 years ago

Read 2 more answers

Find the regular price. Round to the nearest cent . Sales price is 56.78 percent discount is 40%

LenaWriter [7]

$100\%-40\%=60\%\\60\% = \frac{60}{100}=\frac{6}{10}=\frac{3}{5}\\\\$

Numerator is 3 and denominator is 5. So, we divide the sales price by 3 and we multiply by 5 to get the whole fraction, or 100%.

$\frac{56.78*5}{3} \\\frac{283.9}{3} \\94.6333333333333...$

Your regular price sounds recurring, so you might have made a mistake copying.

The answer for what you've got is 94.633333...

5 0

2 years ago

A chemical company makes two brands of antifreeze. The first brand is 20% pure antifreeze, and the second brand is 45% pure anti

Hoochie [10]

The gallons of the first brand of antifreeze should be 60 gallons.

The gallons of the second brand of antifreeze should be 90 gallons.

<h3>What are the linear equations that represent the question?</h3>

0.2f + 0.45s = 52.50 (0.35 x 150) equation 1

f + s = 150 equation 2

Where:

f = gallons of the first antifreeze
s = gallons of the second antifreeze

how many gallons of each brand of antifreeze must be used?

Multiply equation 2 by 0.2

0.2f + 0.2s = 30 equation 3

Subtract equation 3 from equation 2

0.25s = 22.50

s = 22.50 / 0.25

s = 90

f = 150 - 90 = 60 gallons

To learn more about linear functions, please check: brainly.com/question/26434260

#SPJ1

4 0

1 year ago

I need help pls WILL GIVE BRAINLIEST

Alex73 [517]

Answer: (3,-2)

Step-by-step explanation:

$-3x-7y=5\\15x+6y=33$

$5(-3x-7y)=5(5)\\-15x-35y=25$

new system:

$-15x-35y=25\\15x+6y=33$

add.

$-29y=58\\y=-2$

plug in -2 for y

$-3x-7(-2)=5\\-3x+14=5\\-3x=-9\\x=3$

system:

$(3,-2)$

4 0

1 year ago

Read 2 more answers

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago