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3 years ago

EASY 20 points! Please Solve this Algebra 1 question.

Mathematics

2 answers:

Andrei [34K]3 years ago

5 0

Let W = width of the pool

area of back yard = (3W +5)^2

area of pool = W(W+10)

area around pool = (3W+5)^2 - W(W+10)

enyata [817]3 years ago

4 0

length of pool, L = 10 + W
width of pool, W

pool area = W(10+W)

backyard is square shaped.. sides are equal
backyard side length, S = 5 + 3W

Area backyard = (5+3W)(5+3W)
subtract pool area

Area surrounding pool
= (5+3W)(5+3W) - W(10+W)
simplify the expression, foil
= 25 + 30W + 9W^2 - 10W - W^2
combine like terms
A = 8W^2 + 20W + 25

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Sue either travels by bus or walks when she visits the shops. The probability that she catches the bus from the shops is 0.7. Sh

Alina [70]

Corrected Question

Sue travels by bus or walks when she visits the shops. The probability that she catches the bus to the shops is 0.4. The probability she catches the bus from the shops is 0.7. Show the probability that Sue walks at one way is 0.72

Answer:

$P(A \cup B) =0.72$ (Proved)

Step-by-step explanation:

Sue travels by bus or walks when she visits the shops.

Let the event that she catches the bus to the shop=A

Let the event that she catches the bus from the shop=B

P(A)=0.4

P(B)=0.7

Both A and B are independent events.

Therefore,Probability that she catches the bus to and from the shop:

P(A∩B) = 0.4 X 0.7= 0.28

Probability Sue walks at least one way $P(A \cup B) = 1 - P(A \cap B)$

$= 1 - 0.28\\= 0.72$

Hence, the probability that Sue walks at least one way is 0.72.

8 0

3 years ago

Find The number to the angle QPR

butalik [34]

Step-by-step explanation:

you know that QPS adds up to 47

so the sum of RPS and QPR shoukd = 47

(3x - 38) + (7x - 95) = 47

you then solve for x then you can use that to find QPR

8 0

3 years ago

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.

7 0

3 years ago

Wha is the greatest common factor of 1 and 27

Sauron [17]

1. 1’s only factor IS 1, so it’s kind of your only choice.

3 0

3 years ago

Given triangle MNK below, find sin M/ Cos K

meriva

Answer:

The Answer is A

Step-by-step explanation:

its A I got it right

8 0

3 years ago