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Damm [24]
2 years ago
6

Which ordered pair is the solution to the system of equations?

Mathematics
1 answer:
d1i1m1o1n [39]2 years ago
4 0
\begin{cases}-3x+4y=-20\\y=x-4\end{cases}\\\\\\-3x+4(x-4)=-20\\\\-3x+4x-16=-20\\\\x-16=-20\quad|+16\\\\x-16+16=-20+16\\\\\boxed{x=-4}\\\\\\y=x-4\\\\y=-4-4\\\\\boxed{y=-8}

Answer C.
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Bumek [7]

Answer:

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Step-by-step explanation:

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3 years ago
As a fundraiser, the Music Boosters wrapped gifts at the mall one weekend. Out of the 100 gifts they wrapped, 3/10 were for wedd
dangina [55]

Answer: The fraction of the gifts were for birthdays = \dfrac{33}{100}.

Step-by-step explanation:

Given: fraction of gifts for weddings = \dfrac{3}{10}

fraction of gifts for baby showers = \dfrac{37}{100}

Rest of the gifts = 1-\dfrac3{10}-\dfrac{37}{100}

=\dfrac{100-30-37}{100}\ \ \ [\text{Making like fractions}]\\\\=\dfrac{100-67}{100}\\\\=\dfrac{33}{100}

Hence, the fraction of the gifts were for birthdays = \dfrac{33}{100}.

4 0
2 years ago
John buys 3 pounds of cherries and 2 pounds of bananas he pays a total of $24.95 the bananas cost $6.50 less per pound than the
Murrr4er [49]

John will pay $8.68 for the combined cost of 1 pound of banana and 1 pound of cherries.

Let: b=cost of banana per pound and c=cost of cherries per pound

Equation 1: For 3 pounds of cherries and 2 pounds of bananas, John pays a total of $24.95.

3c + 2b =$24.95

Equation 2: The cost of bananas is $6.50 less than a pound of cherries.

b= c - $6.50

We can substitute the second equation into the first one to solve for the cost of cherries per pound.

3c + (2)(c-$6.50)= $24.95

3c + 2c -$13.00 = $24.95

5c = $24.95 + $13.00

c = $7.59

Substituting the value of c to the second equation to solve for b.

b= $7.59 - $6.50 = $1.09

The combined cost of 1 pound of banana and 1 pound of cherries is $1.09 + $7.59 or $8.68.

For more information regarding the system of equations, please refer to brainly.com/question/25976025.

#SPJ4

4 0
1 year ago
What do you set each term equal to in order to solve for all of the "roots" of a function? No links please :)​
Citrus2011 [14]

Answer:

Solving quadratic equations can be difficult, but luckily there are several different methods that we can use depending on what type of quadratic that we are trying to solve. The four methods of solving a quadratic equation are factoring, using the square roots, completing the square and the quadratic formula.

Step-by-step explanation:

can you mark me brainliest

3 0
2 years ago
Leo the Rabbit
Alla [95]

Answer:

2?

Step-by-step explanation: because he can hop them one step at a time or two steps at a time

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