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3 years ago

A student council sells 44 bouquets for $176. Which equation represents the

Mathematics

1 answer:

vesna_86 [32]3 years ago

4 0

Answer:

$y = 4x$

Step-by-step explanation:

Given

$x = 44$

$y = 176$

Required

Determine the relationship between x and y

First, we calculate the unit rate, using:

$y = kx$

Where k is the unit rate.

Substitute 176 for y and 44 for y

$176 = k * 44$

Divide both sides by 44

$\frac{176}{44} = \frac{44*k}{44}$

$\frac{176}{44} = k$

$4 = k$

$k = 4$

Substitute 4 for k in $y = kx$

$y = 4x$

Hence, the relationship between x and y is: $y = 4x$

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the seventh grade class is putting on a variety show to raise money. it cost 700 dollars to rent the banquet hall that they are

nexus9112 [7]

Answer:

Since the question is partial, i will assume 2 scenarios:

They need to raise 1000 income

They need to make 1000 as profit

If $1000 as income:

Each ticket costs $15, so tickets would bring them $1000 income. Fractional ticket is not possible, so rounding gives us 67 tickets as the answer.

If $1000 as profit:

Their cost of renting is $700. We know that .

So, . So, to raise $1700, we need tickets. Fractional ticket is not possible, so rounding gives us 114 tickets as the answer.

ANSWER:

If need to raise atleast $1000 as income, they need to sell 67 tickets.

If need to raise atleast $1000 as profit, they need to sell 114 tickets.

The most probable answer would be 114 tickets

PLZ MARK ME BRAINLIEST

4 0

3 years ago

Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)

Leni [432]

Alrighty

remember some rules
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}=\frac{1}{x^m}$
and
$x^0=1$ for all real values of x
and
$(a^b)^c=a^{bc}$
and
$(\frac{a}{b})^c=\frac{a^c}{b^c}$
and
(a/b)/(c/d)=(ad)/(bc)
and
$(a^b)(a^c)=a^{b+c}$
and
$\frac{a^m}{a^n}=a^{m-n}$
and don't forget pemdas
example: $-x^m=-1(x^m)$ but $(-x)^m=(-1)^m(x^m)$

so

4.
$(\frac{c^{-2}}{2})^{-2}=$

$\frac{(c^{-2})^{-2}}{2^{-2}}=$

$\frac{c^4}{\frac{1}{2^2}}=$

$\frac{c^4}{\frac{1}{4}}=$

$4c^4$

5.

$\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=$

$\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=$

$\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=$

$\frac{(a^4bc^5)b^3}{(-1)(a^2)}=$

$\frac{-a^4b^4c^5}{a^2}=$

$-a^2b^4c^5$

3 0

3 years ago

Consider triangle ABC where AB=X+5, BC=X-2, area is 30cm squared. Find X by solving X^2+3X-70=0, and find the perimeter.

klasskru [66]

Answer:

What the hell AC is not given so you can't find the Perimeter

Step-by-step explanation:

x^2+3x-70=0

(x-7)(x+10)=0

x-7=0

x+10=0

x=-10

AB=x+5

AB=12

AB=-5

BC=x-2

BC=5

BC=-12

Area of ABC = 30cm2

Perimeter of ABC = AB+BC+AC

8 0

3 years ago

I don't understand this question can you help me please

Maslowich

I believe the answer is A because the formula for area is length times width.

5 0

3 years ago

Read 2 more answers

The mid-point or number which divides a series of values into two groups of equal numbers of values is referred to as the:

Over [174]

The mid-point or the number that divides a series of values into two groups of equal numbers of values is referred to as the median. To get the median value, first sort the numbers from highest to lowest or lowest to highest and then pick the middle number. However, for even numbered sets of numbers, take the average of the two middle numbers and the resulting number is the median.

3 0

3 years ago