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4 years ago

Simplify the expression

Mathematics

2 answers:

11111nata11111 [884]4 years ago

8 0

This can be rewritten as 8.2 - 5.6 so it is easier to solve.

8.2 - 5.6 = 2.6

Hope this helps!

Alla [95]4 years ago

4 0

<span>–5.6 + 8.2

The answer is 2.6 because with a - and + you </span>subtract the two number and whichever is the highest number that is the sign that you put. In this case, the highest number is a +8.2 so your answer will be POSITIVE.

If the highest number was a negative for example, 5.6+ -8.2 then your answer will be -2.6

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Answer:

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1 year ago

Example:The missing value is x than Sin 7 / 18

Darya [45]

Answer:

44.9

Step-by-step explanation:

12/17 = PQ/PR

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3 years ago

Read 2 more answers

Jason jumped off of a cliff into the ocean in Acapulco while vacationing with some friends. His height above ocean measured in f

lbvjy [14]

Answer:

480 feet.

Step-by-step explanation:

We are told that the function $h(t)=-16t^2+16t+480$ models Jason's height above ocean measured in feet as a function of time and t is the time in seconds from jumping off.

To find the height of cliff we need to substitute t=0 in our given function as at t=0 we will get Jason's height above ocean which is same as the height of the cliff.

Upon substituting t=0 in our function we will get,

$h(0)=-16(0)^2+16*0+480$

$h(0)=-0+0+480$

$h(0)=480$

Since, the function gives Jason's height above ocean in feet, therefore, the cliff was 480 feet high.

8 0

3 years ago

Use the definition of continuity to determine whether f is continuous at a.

dmitriy555 [2]

$f(x)$ will be continuous at $x=a=7$ if
(i) $\displaystyle\lim_{x\to7}f(x)$ exists,
(ii) $f(7)$ exists, and
(iii) $\displaystyle\lim_{x\to7}f(x)=f(7)$ .

The second condition is immediate, since $f(7)=8918$ has a finite value. The other two conditions can be established by proving that the limit of the function as $x\to7$ is indeed the value of $f(7)$ . That is, we must prove that for any $\varepsilon>0$ , we can find $\delta>0$ such that

$|x-7|$

Now,

$|f(x)-f(7)|=|5x^4-9x^3+x-8925|$

Notice that when $x=7$ , we have $5x^4-9x^3+x-8925=0$ . By the polynomial remainder theorem, we know that $x-7$ is then a factor of this polynomial. Indeed, we can write

$|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|$

This is the quantity that we do not want exceeding $\varepsilon$ . Suppose we focus our attention on small values $\delta$ . For instance, say we restrict $\delta$ to be no larger than 1, i.e. $\delta\le1$ . Under this condition, we have

$|x-7|$

Now, by the triangle inequality,

$|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275$

If $|x|$ , then this quantity is moreover bounded such that

$|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955$

To recap, fixing $\delta\le1$ would force $|x|$ , which makes

$|x-7||5x^3+26x^2+182x+1275|$

and we want this quantity to be smaller than $\varepsilon$ , so

$6955|x-7|$

which suggests that we could set $\delta=\dfrac{\varepsilon}{6955}$ . But if $\varepsilon$ is given such that the above inequality fails for $\delta=\dfrac{\varepsilon}{6955}$ , then we can always fall back on $\delta=1$ , for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set $\delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}$ .

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that $f(x)$ is indeed continuous at $x=7$ .

5 0

3 years ago

The ratio of quarters to dimes in a coin collection is 5:3. You add the same number of new quarters as dimes

Anon25 [30]

The ratio of quarters to dimes is not still 5 : 3

<u>Solution:</u>

Given that ratio of quarters to dimes in a coin collection is 5:3 .

You add same number of new quarters as dimes to the collection .

Need to check if ratio of quarters to dimes is still 5 : 3

As ratio of dimes and quarters is 5 : 3

lets assume initially number of quarters = 5x and number of dimes = 3x.

Now add same number of new quarters as dimes to the collection

Let add "x" number of quarters and "x" number of dimes

So After adding,

Number of quarters = initially number of quarters + added number of quarters = 5x + x = 6x

Number of dimes = initially number of dimes + added number of dimes

= 3x + x = 4x

New ratio of quarters to dimes is 6x : 4x = 3 : 2

So we have seen here ratio get change when same number of new quarters and dimes is added to the collection

Ratio get change from 5 : 3 when same number of new quarters and dimes is added to the collection and new ratio will depend on number of quarters and dimes added to collection.

3 0

4 years ago