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3 years ago

Geometry!!!!!!!!!!!!!!!

Mathematics

1 answer:

Nimfa-mama [501]3 years ago

6 0

Answer:

x=4

Step-by-step explanation:

1. 7x+56=14x+28 using distributive property (with the number of stes you need to have, this may count as 2)

2. 56=7x+28 by subtracting 7x from each side

3. 28=7x by subtracting 28 from both sides

4. x=4 by dividng by 7

5. plug back into equation

7(4)+56=14(4+2)

28+56=14(6)

84=84

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Can someone explain this to me?<br><br> Please solve step by step. I don't know how to do this.

ololo11 [35]

Given,

y = -3x+1 (EQ1)
y = 2x-4 (EQ2)

Equate EQ1 and EQ2

EQ1 = EQ2
-3x+1 = 2x-4
-3x-2x = -4-1
x = 1

y = -3x+1
y = -3(1)+1
y = -3+1
y = -2

(1, -2)
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6 0

3 years ago

Please help me out……..

Stells [14]

Answer:

y = 20

Step-by-step explanation:

y = kx

24 = 6k

24/6 = 6k/6

k = 4

y = kx

y = 4(5)

y = 20

3 0

3 years ago

Read 2 more answers

Find the value of x and y

elena55 [62]

Answer:

x=50 and y=20

Step-by-step explanation:

I hope this helped!

3 0

2 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

Determine the volume of the square pyramid below h-4 b1-6 b2-6 144 square units 36 square units 72 cubic units 48 cubic units

bagirrra123 [75]

Answer:

V = 48 cubic units

Step-by-step explanation:

$V=b^{2} \frac{h}{3}$ Use this equation to find the volume

$V=6^{2} \frac{4}{3}$ Simplify the exponent

$V=36(\frac{4}{3})$ Multiply

V = 48 cubic units

If this answer is correct, please make me Brainliest!

4 0

3 years ago