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3 years ago

10

How would I solve this?

"\frac{2\sqrt{21} }{12}" alt="\frac{2\sqrt{21} }{12}" align="absmiddle" class="latex-formula">

1 answer:

egoroff_w [7]3 years ago

3 0

U just solve 2 square root 21 and then divide by 12

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The axis of symmetry for a function in the form f(x) = x2 + 4x − 5 is x = −2. What are the coordinates of the vertex of the grap

Studentka2010 [4]

Answer:

D) $(-2,-9)$

Step-by-step explanation:

<u>Vertex Form of a Vertical Parabola:</u>

$y=a(x-h)^2+k$

Vertex -> $(h,k)$

Axis of Symmetry -> $x=h$

Vertical Scale Factor -> $a$

To turn $f(x)=x^2+4x-5$ into vertex form, we need to complete the square on the right side
Therefore, if $f(x)=0$ , then $0+9=x^2+4x-5+9$ completes the square on the right side
This becomes $9=(x+2)^2$
This means that our function in vertex form is $f(x)=(x+2)^2-9$

Therefore, the vertex of the graph is $(h,k)=(-2,-9)$ .

3 0

3 years ago

An object with a mass of 3.0 g raises the level of water in a graduated cylinder from 25.1 mL to 33.1 mL. What is the density of

evablogger [386]

The displaced volume of the water, because of the submersion of the object, is equal to the volume of the object.

So to find the volume of the object we calculate the difference in volume of water:
33.1-25.1 = 8 ml

Since 1 mL of water has the volume of 1cm^3, then the volume of the object is 8 cm^3.

The density of the object is found by the formula:

density= mass/volume= (3 g) / (8 cm^3) = 3/8 (g/cm^3)= 0.375 g/cm^3

Answer: 0.375 g/cm^3

7 0

4 years ago

Let X, the number of flaws on the surface of a randomly selected boiler of a certain type, have a Poisson distribution with para

kvasek [131]

Answer:

(a) 0.932

(b) 0.0653

(c) 0.032

(d) 0.316

(e) 0.251

Step-by-step explanation:

From the table with mean parameter μ = 5, we can compute the following cumulative and density probability

(a) $P(X \leq 8) = 0.932$ (cumulative)

(b) P(X = 8) = 0.0653 (density)

(c) $P(9 \leq X) = 1 - P(X \leq 9) = 1 - 0.968 = 0.032$ (cumulative)

(d) $P(5 \leq X \leq 8) = P(X \leq 8) - P(X \leq 5) = 0.932 - 0.616 = 0.316$ (cumulative)

(e) $P(5 < X < 8) = P(X \leq 8) - P(X \leq 5) - P(X = 8) = 0.932 - 0.616 - 0.0653 = 0.251$

5 0

3 years ago

3 and 1 half to the power of 2 simplify

Elenna [48]

3 and 1 half is also known as 3.5,
3.5 x 3.5 = 12.25
in fractions that is 12 and 1 quarter

8 0

3 years ago

Read 2 more answers

. 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