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tangare [24]
3 years ago
10

What do you use to remove any grouping symbols in math

Mathematics
1 answer:
QveST [7]3 years ago
6 0
<span>When we remove those parentheses, the sign of every term
within the parentheses changes.</span>

Problem 2.   Remove the parentheses.

a)   p + (q −r + s) <span>= p + q − r + s</span>

b)   p − (q −r + s) <span>= p − q + r − s</span>

In each of the following problems, remove the parentheses, then simplify
by adding the numbers.

For example,

<span><span><span>(x − 3) − (y − 4)</span>  =  <span>x − 3 − y + 4</span></span> <span>   =  <span>x − y + 1.</span></span></span>

The sign preceding  (x − 3)  is understood to be + . Therefore the signs within those parentheses do not change.

But the sign preceding (y − 4) is minus.  Therefore, y changes to −y, and −4 changes to +4.

Finally, it is the style in algebra to to write the literal terms, x − y, to the left of the numerical term.

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You get 17 out of 22 correct on a math test what is your percent of correct answers
Zinaida [17]

Answer:

77%

Step-by-step explanation:

divide 17 by 22

6 0
3 years ago
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PLEASE HELP ME I NEED THE ANSWER ASAP THANK YOU SO MUCH!!
Ostrovityanka [42]

dark blue is the answer

6 0
3 years ago
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Decide whether the relation is a function?
Nana76 [90]

Answer:

A property of a function is that it gives one output for an input

if a function gives one output for an input, it is a function and otherwise it is not a function otherwise

here, we can see that we have 2 outputs for the input '6'. therefore, it is NOT a function

3 0
3 years ago
Please solve the problem with steps
Debora [2.8K]

Answer:

Infinite series equals 4/5

Step-by-step explanation:

Notice that the series can be written as a combination of two geometric series, that can be found independently:

\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}

The first one: (\frac{1}{2})^{n-1} is a geometric sequence of first term (a_1) "1" and common ratio (r) " \frac{1}{2} ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2

The second one: \frac{1}{6^{n-1}} is a geometric sequence of first term "1", and common ratio (r) " \frac{1}{6} ". Again, since the common ratio is smaller than one, we can find its infinite sum:

Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{6} } =\frac{1}{\frac{5}{6} } =\frac{6}{5}

now we simply combine the results making sure we do the indicated difference: Infinite total sum= 2-\frac{6}{5} =\frac{10-6}{5} =\frac{4}{5}

8 0
3 years ago
Read 2 more answers
F(x) = 1/(5x-6)Find the derivative of this function
siniylev [52]

Given

f(x)=\frac{1}{5x-6}

To find the derivative of the give function,we apply below exponent rule first:

\frac{1}{a}=a^{-1}

So,

Next, we apply the chain rule:

So,

We substitute back u=(5x-6):

Simplify:

Therefore, the answer is:

=-\frac{5}{(5x-6)^2}

5 0
1 year ago
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