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4 years ago

Please help! Given the function problem

Mathematics

2 answers:

dedylja [7]4 years ago

7 0

Answer:

Step-by-step explanation:

So we have the function:

$f(x)=\frac{9}{5}x+32$

To find the inverse, flip f(x) and x, change f(x) to f⁻¹(x), and solve for it. Thus:

$x=\frac{9}{5}f^{-1}(x)+32$

Subtract 32 from both sides:

$x-32=\frac{9}{5}f^{-1}(x)$

Multiply both sides by 5/9. The right side will cancel. Thus:

$f^{-1}(x)=\frac{5}{9}(x-32)$

Our answer is B.

And we're done!

evablogger [386]4 years ago

6 0

Answer:

I think A!

If not I'm srry! But i think its A! :)

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3 years ago

To prove that DEF congruent DGF by SAS,what additional information is needed

lisabon 2012 [21]

Answer:

∠DFE ≅ ∠DFG

Step-by-step explanation:

we know that

Triangles are congruent by SAS, if any pair of corresponding sides and their included angles are equal in both triangles

In this problem we have that

The pair of corresponding sides that are equal are

EF=GF and DF (is the same side for both triangles)

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8 0

4 years ago

Read 2 more answers

L :V --> W is a linear transformation. Prove each of the following (a) ker L is a subspace of V. (b) range L is a subspace of

iragen [17]

Answer:

a) Assume that $x,y\in\ker L$ , and $\alpha$ is a scalar (a real or complex number).

First. Let us prove that $\ker L$ is not empty. This is easy because $L(0_V)=0_W$ , by linearity. Here, $0_V$ stands for the zero vector of V, and $0_W$ stands for the zero vector of W.

Second. Let us prove that $\alpha x\in\ker L$ . By linearity

$L(\alpha x) = \alpha L(x)=\alpha 0_W=0_W$ .

Then, $\alpha x\in\ker L$ .

Third. Let us prove that $y+ x\in\ker L$ . Again, by linearity

$L(x+y)=L(x)+L(y) = 0_W + 0_W=0_W$ .

And the statement readily follows.

b) Assume that $u$ and $v$ are in range of $L$ . Then, there exist $x,y\in V$ such that $L(x)=u$ and $L(y)=v$ .

First. Let us prove that range of $L$ is not empty. This is easy because $L(0_V)=0_W$ , by linearity.

Second. Let us prove that $\alpha u$ is on the range of $L$ .

$\alpha u = \alpha L(x) = L(\alpha x) = L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)=\alpha u$ . Thus $\alpha u$ is in the range of $L$ .

Third. Let us prove that $u+v$ is in the range of $L$ .

$u+v = L(x)+L(y) = L(x+y)=L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)= u +v$ . Thus $u +v$ is in the range of $L$ .

Notice that in this second part of the problem we used the linearity in the reverse order, compared with the first part of the exercise.

Step-by-step explanation:

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