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Lynna [10]
3 years ago
10

Use distributive property to rewrite each algebraic expression: 8(s + 5)

Mathematics
2 answers:
pishuonlain [190]3 years ago
7 0

Answer: The correct answer is 8s+40

Step-by-step explanation: If it's correct can i get brainliest..... hopes this helps.

igor_vitrenko [27]3 years ago
4 0

Answer:

8s + 40

Step-by-step explanation:

8*s = 8s

8*s = 40

8s + 40

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How to find only f on a graph that defines a function
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Answer:

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Step-by-step explanation:

we can represent a function using a graph. Graphs display many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. We typically construct graphs with the input values along the horizontal axis and the output values along the vertical axis.

The vertical line test can be used to determine whether a graph represents a function. A vertical line includes all points with a particular

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value. The

y

value of a point where a vertical line intersects a graph represents an output for that input

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value. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that

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value has more than one output. A function has only one output value for each input value.

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3 years ago
Divide. Give the quotient and remainder.<br> 33 - 8<br> :))))
Ilya [14]
Answer: 9 remainder 2

Explanation: 9 is the closest 6 can get to 56 without going over. Since 9•6=54, and 56-54=2 , 2 is the remainder.
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3 years ago
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15 POINTS ITS MY TEST ALREADY MISSING PLEASE HELP ON THIS ONE
const2013 [10]
Melanie walked 2.2 miles. Cathy walked 1.75 times as far. This means that you need to do 2.2 * 1.75 which equals 3.85. To find how much further, do 3.85 - 2.2 which equals 1.65. Therefore Cathy walked 1.65 more miles than Melanie.
3 0
2 years ago
. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,
Eva8 [605]

Answer:

Elements are of the form

 (i) [a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}

(ii) [a,b)=\{[x,y) :a\leq x

(iii)(a,a]=\{(x,y] :a

(iv)(a,a)=\{(x,y): a

(v) (a,b) where a>b=\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}

(vi) [a,b] where a>b=\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi)  [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........

Because there is no increment so if a\in \mathbb R then the set  (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

(a,b)=\{(5,0),(5,1),(5,2).....\} e.t.c

So this set is connected and we know singletons are connected in \mathbb R. Hence given set is empty.

(vi) [a,b] where a\leq b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

[a,b]=\{[5,0],[5,1],[5,2].....\} e.t.c

So this set is connected and we know singletons are connected in \mathbb R. Hence given set is empty.

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olga nikolaevna [1]

Answer:

2nd point

Step-by-step explanation:

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3 years ago
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