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3 years ago

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A man has a 10m X 10m square garden. In the center is a 2m X 2m square patch which he cannot use. He divides his usable space in

to four congruent rectangular patches, each of which measures

1 answer:

ra1l [238]3 years ago

8 0

Answer:

4m X 6m

Step-by-step explanation:

This is because if there is a 2m square in the middle, there is 8 m of usable space left along both sides of the garden.

Because the 2m square was in the middle, there is 8/2 = 4m along each width of the 4 small rectangles.

Becuase 4m is the width, there is 10 - 4 = 6m along each rectangle's length.

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Heyy pls help! worth 20 points, and please only help if you know the answer. also no links those r kinda annoying lol

iragen [17]

Step 1: Find the slope:

$m= \dfrac{y_2-y_1}{x_2-x_1}= \dfrac{-2-7}{4-(-2)} = \dfrac{-9}{6}= -\dfrac{3}{2}$

This gives you $y=-\dfrac{3}{2}x+b$ , but we need to find b.

To find b, substitute in one (x,y) pair and it doesn't matter which one. I'll go with (4,-2):

$\begin{aligned}-2&=-\dfrac{3}{2}(4)+b\\[0.5em]-2&=-6+b\\[0.5em]4&=b\end{aligned}$

Now take that b-value and plug in into the slope-intercept form:

$y=-\dfrac{3}{2}x+4$

It's always a good idea to toss in the other x-value from the other point, to make sure it checks out.

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How do you write 50% as a fraction?

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Read 2 more answers

Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp

shtirl [24]

Answer:

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If $\phi: G \longrightarrow G$ is an homomorphism, then it must hold

that $b^{-1}a^{-1}=(ab)^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1}$ ,

but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

$\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)$

which tells us that $\phi$ is a homorphism. The kernel of $\phi$

is the set of elements g in G such that $g^{n}=1$ . However,

$\phi$ is not necessarily 1-1 or onto, if $G=\mathbb{Z}_6$ and

n=3, we have

$kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}$

(c) If $z_1,z_2 \in \mathbb{C}^{\times}$ remeber that

$|z_1 \cdot z_2|=|z_1|\cdot|z_2|$ , which tells us that $\phi$ is a

homomorphism. In this case

$kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}$ , if we write a

complex number as $z=x+iy$ , then $|z|=x^2+y^2$ , which tells

us that $kern(\phi)$ is the unit circle. Moreover, since

$kern(\phi) \neq \{1\}$ the mapping is not 1-1, also if we take a negative

real number, it is not in the image of $\phi$ , which tells us that

$\phi$ is not surjective.

(d) Remember that $e^{ix}=\cos(x)+i\sin(x)$ , using this, it holds that

$\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)$

which tells us that $\phi$ is a homomorphism. By computing we see

that $kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \}$ and

$Im(\phi)$ is the unit circle, hence $\phi$ is neither injective nor

surjective.

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3 years ago