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2 years ago

What is squre root of 8

Mathematics

1 answer:

IceJOKER [234]2 years ago

6 0

Answer:

2.838

Step-by-step explanation:

put it in the calculator.

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HELP WILL MARK YOU BRAINLIEST!!!

Vlad [161]

Answer:

<PQR=132°

plz mark me brainliest

3 0

3 years ago

Please help simple pre algebra!!!

xxMikexx [17]

C=pi(d)

So d=7ft

Then pi(7ft)=21.98

3 0

3 years ago

Read 2 more answers

Assume {v1, . . . , vn} is a basis of a vector space V , and T : V ------> W is an isomorphism where W is another vector spac

Degger [83]

Answer:

Step-by-step explanation:

To prove that $w_1,\dots w_n$ form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

Recall that the set $w_1,\dots w_n$ is linearly independent if and only if the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$ implies that

$\lambda_1 = \cdots = \lambda_n$ .

Recall that $w_i = T(v_i)$ for i=1,...,n. Consider $T^{-1}$ to be the inverse transformation of T. Consider the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$

If we apply $T^{-1}$ to this equation, then, we get

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0$

Since T is linear, its inverse is also linear, hence

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots + \lambda_nT^{-1}(w_n)=0$

which is equivalent to the equation

$\lambda_1v_1+\dots + \lambda_nv_n =0$

Since $v_1,\dots,v_n$ are linearly independt, this implies that $\lambda_1=\dots \lambda_n =0$ , so the set $\{w_1, \dots, w_n\}$ is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist $a_1, \dots a_n$ such that

$w = a_1w_1+\dots+a_nw_n$

Since T is surjective, there exists a vector v in V such that T(v) = w. Since $v_1,\dots, v_n$ is a basis of v, there exist $a_1,\dots a_n$ , such that

$a_1v_1+\dots a_nv_n=v$

Then, applying T on both sides, we have that

$T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w$

which proves that $w_1,\dots w_n$ generate the whole space W. Hence, the set $\{w_1, \dots, w_n\}$ is a basis of W.

Consider the linear transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of $\mathbb{R}^2$ given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of $\mathbb{R}^2$

8 0

3 years ago

A teacher plans to buy air fresheners that cost $3.50 each. If the teacher can spend no more than $10 on the air fresheners and

iris [78.8K]

Answer:

Step-by-step explanation:

We need to create an equation to represent how many air fresheners the teacher can buy.

Teacher has $10 and a $3 discount coupon which would give her an extra $3 to spend. Each air freshener is $3.50.

x = number of air fresheners

($10 + $3) / $3.50 = x can be written as $13/$3.50 = x

We add $10 and $3 because that gives us the total amount she has to spend.

We divide by $3.50 because that is the cost of each air freshener. The result is the number of air fresheners she can buy.

In this case.

$13/$3.50 = x

3.71 = x BUT we can't buy part of an air freshener, so she can purchase 3 air fresheners.

8 0

3 years ago

Find the average rate of change of the function over the given interval

sattari [20]

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{4}\\
t_2=\frac{3\pi }{4}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{4} \right)-h\left( \frac{\pi }{4} \right)}{\frac{3\pi }{4}-\frac{\pi }{4}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{4} \right)}{sin\left( \frac{3\pi }{4} \right)}-\frac{cos\left( \frac{\pi }{4} \right)}{sin\left( \frac{\pi }{4} \right)}}{\frac{\pi }{2}}\implies \cfrac{-1-1}{\frac{\pi }{2}}\implies \cfrac{-2}{\frac{\pi }{2}}\implies -\cfrac{4}{\pi }\\\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{3}\\
t_2=\frac{3\pi }{2}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{2} \right)-h\left( \frac{\pi }{3} \right)}{\frac{3\pi }{2}-\frac{\pi }{3}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{2} \right)}{sin\left( \frac{3\pi }{2} \right)}-\frac{cos\left( \frac{\pi }{3} \right)}{sin\left( \frac{\pi }{3} \right)}}{\frac{9\pi -2\pi }{6}}\implies \cfrac{\frac{0}{-1}-\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}{\frac{7\pi }{6}}\implies\cfrac{-\frac{1}{\sqrt{3}}}{\frac{7\pi }{6}}\implies -\cfrac{\sqrt{3}}{3}\cdot \cfrac{6}{7\pi }
\\\\\\
-\cfrac{2\sqrt{3}}{7\pi }$

8 0

3 years ago