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gregori [183]
3 years ago
5

Jylene and her brother had a total of $35.19 to buy school supplies. Jylene’s school supplies cost $12.78

Mathematics
2 answers:
Allushta [10]3 years ago
7 0

Answer:

whats the question

Step-by-step explanation:

Anni [7]3 years ago
6 0

Answer:

What is the question???

Step-by-step explanation:

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A path 1 m wide is build around a garden that is a 30-meter square. What is the area taken by the path?
maria [59]

Length of outer boundary =30m

So area of outer boundary =30×30

= 900m²

Length of inner boundary of path=30-1= 29m

Inner area =29×29 =841m²

So area of path is= outer area - inner area

=900- 841

= 59m²

6 0
2 years ago
ASAP!! I need help. -w-Which answer choice shows 5 +0.9 + 0.06 written in standard form?
miss Akunina [59]
The answer is c . 5.96
7 0
3 years ago
Read 2 more answers
Can someone please help me with comparing function
Minchanka [31]
Understand that a function<span> is a rule that assigns to each input exactly one output. The graph of a </span>function<span> is the set of ordered pairs consisting of an input and the corresponding output.</span>
8 0
3 years ago
Given that f(x)=x2-1 and g(x)=2x+8 find (g-f)(10)
irga5000 [103]

Answer:

<h2>(g-f)(10) = - 71</h2>

Step-by-step explanation:

f(x) = x² - 1

g(x) = 2x + 8

To find (g-f)(10) first find ( g - f)(x)

To find ( g - f)(x) subtract f(x) from g(x)

That's

( g - f)(x) = 2x + 8 - ( x² - 1)

Remove the bracket

( g - f)(x) = 2x + 8 - x² + 1

Simplify

( g - f)(x) = - x² + 2x + 9

To find (g-f)(10) substitute the value in the bracket that's 10 into ( g - f)(x)

That is

(g-f)(10) = -(10)² + 2(10) + 9

= - 100 + 20 + 9

= - 100 + 29

= - 71

Hope this helps you

7 0
3 years ago
Let f (x) = 3x − 1 and ε &gt; 0. Find a δ &gt; 0 such that 0 &lt; ∣x − 5∣ &lt; δ implies ∣f (x) − 14∣ &lt; ε. (Find the largest
s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and \varepsilon>0, we want to find a \delta such that |f(x)-14|

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that |x-5|. Then

|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta.

Then, if |x-5|, then |f(x)-14|. So, in this case, if 3\delta \leq \varepsilon we get that |f(x)-14|. The maximum value of delta is \frac{\varepsilon}{3}.

By definition, this procedure proves that \lim_{x\to 5}f(x) = 14. Note that f(5)=14, so this proves that f is continous at x=5.

3 0
2 years ago
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