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Sergio [31]
3 years ago
6

6p - 5 = 13 What equals p?

Mathematics
2 answers:
Alex73 [517]3 years ago
4 0

Answer:

p = 3

Step-by-step explanation:

6p - 5 = 13

Add 5 to both sides to eliminate -5

6p - 5 + 5 = 13 + 5

6p = 18

Divide both sides by 6 to isolate p

6p/6 = 18/6

p = 3

Andreas93 [3]3 years ago
3 0

Answer:

6p - 5 + 13   P = 3

Step-by-step explanation:

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Fofino [41]
The Measure of the Missing Angles can be found by this formula: x+y+z= 180°.
You already know the measure of 1 Angle, which is 30°, right?
You also know that this Triangle is a Right Triangle, so the Square for One Angle indicates that the Angle is 90°.
y= 90°, and z= 30°, and you know that the Total Measure of any Triangle is 180° Total.
90°+30° = 120°, and 180°-120°= 60°, so finally, x= 60°, and y=90°, and z= 30°.
4 0
3 years ago
Write A number d minus 2 is less than -1 as an inequality
laiz [17]

Answer:

d-2 < -1

Step-by-step explanation:

4 0
3 years ago
A number decreased by 23% equals 123
Elden [556K]

Answer:

100 is inital value of that decrease

7 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.

8 0
3 years ago
State the domain restriction(s) in interval notation of \displaystyle f\left(g\left(x\right)\right)f(g(x)) given: \displaystyle
DENIUS [597]

Answer:

The interval notation for the domain is [\frac{23}{3},\infty  ].

Step-by-step explanation:

Consider the provided information.

It is given that \:f\left(x\right)=\sqrt{3x-2},\:\text{ and }\:g\left(x\right)=x-7

We need to find the value of f\left(g\left(x\right)\right).

Put the value of g(x) in  f\left(g\left(x\right)\right).

f\left(g\left(x\right)\right)=f(x-7)  ....(1)

Now, put x=x-7 in \:f\left(x\right)=\sqrt{3x-2}

\:f\left(x-7\right)= \sqrt{3(x-7)-2}

\:f\left(x-7\right)= \sqrt{3x-21-2}

\:f\left(x-7\right)= \sqrt{3x-23}

From equation 1.

f\left(g\left(x\right)\right)=\:f\left(x-7\right)= \sqrt{3x-23}

The domain of the function is the set of input values for which a function is defined.

Here, the value of 3x-23 should be greater or equal to 0 as the square root of a negative number is not real.

Domain= 3x-23\geq0

x\geq\frac{23}{3}

The value of x is all real number greater than \frac{23}{3}.

Hence, the interval notation for the domain is [\frac{23}{3},\infty  ].

7 0
2 years ago
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