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3 years ago

What is the general process for solving an equation with one variable

Mathematics

2 answers:

Mice21 [21]3 years ago

5 0

The general process is by grouping like term I. E the unknown to the left hand side while value to the right hand side

ivanzaharov [21]3 years ago

4 0

Thanks for the question!

First, make sure all the variable are on one side and all the numbers are on the other. Then, combine like terms. If necessary, isolate the variable.

Hope this helps!

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Y=4/3x-3. Slope is 4/3

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3 years ago

Alex and Leonardo purchase to matinee movie tickets mantiee ticket cost $6.50 and drinks cost $5.50 and a bag of popcorn cost $6

Veseljchak [2.6K]

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3 years ago

A newspaper and three hot chocolates cost $7. Two newspapers and two hot chocolates cost $6. How much does one hot chocolate cos

pochemuha

Answer:

the cost of one hot chocolate is $2.

Step-by-step explanation:

Let x be the hot chocolate cost and y be the newspaper cost.

Given:

A newspaper and three hot chocolates cost $7

So, the equation is

$3x+y=7$ --------(1)

Two newspapers and two hot chocolates cost $6

So, the second equation is.

$2x+2y=6$ ---------(2)

Find one hot chocolate cost?

Solve equation 1 for y.

$3x+y=7$

$y=7-3x$

Put y value in equation 2

$2x+2(7-3x)=6$

$2x+14-6x=6$

$14-4x=6$

$-4x=6-14$

$-4x=-8$

negative sign is cancelled both side.

$x=\frac{8}{4}$

$x=2$

Therefore, the cost of one hot chocolate is $2.

3 0

2 years ago

Multiply. Write your answer in standard form<br> 3i(-5+i)

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2 years ago

A student repeatedly measures the mass of an object using a mechanical balance and gets the following values: 560 g, 562 g, 556

MrRissso [65]

Answer: 2.76 g

Step-by-step explanation:

The formula to find the standard deviation:-

$\sigma=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{n}}$

The given data values : 560 g, 562 g, 556 g, 558 g, 560 g, 556 g, 559 g, 561 g, 565 g, 563 g.

Then, $\overline{x}=\dfrac{\sum_{i=1}^{10} x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{560+562+556+558+560+556+559+561+565+563}{10}\\\\\Rightarrow\ \overline{x}=\dfrac{5600}{10}=560$

Now, $\sum_{i=1}^{10}(x_i-\overline{x})^2=0^2+2^2+(-4)^2+(-2)^2+0^2+(-4)^2+(-1)^2+1^2+5^2+3^2\\\\\Rightarrow\ \sum_{i=1}^{10}(x_i-\overline{x})^2=76$

Then, $\sigma=\sqrt{\dfrac{76}{10}}=\sqrt{7.6}=2.76$

Hence, the standard deviation of his measurements = 2.76 g

6 0

3 years ago