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attashe74 [19]
3 years ago
9

I'm sorry please help.

Mathematics
2 answers:
Dafna11 [192]3 years ago
8 0

Answer:

x= 5

y= -4

Step-by-step explanation:

x= 17+3y so you substitute what you got for x in so

17 +3y +7y = -23

10y=-40

y=-4 now you put -4 in for y in the first equation

x = 17 + (3 times -4)

x = 17 - 12

x=5

Katena32 [7]3 years ago
3 0

Answer:

-4y=40

y=-10

x-3(-10)=17

x-30=17

x=17+30

x=47

Step-by-step explanation:

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Ratling [72]

Answer:

B IS THE ANSWER

Step-by-step explanation:

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3 years ago
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For f(x)=x/4-3 and g(x)=4x^2+2x-4 find (f+g)(x)
blondinia [14]

f(x)=\dfrac{x}{4}-3=\dfrac{1}{4}x-3\\\\g(x)=4x^2+2x-4\\\\(f+g)(x)=f(x)+g(x)=\dfrac{1}{4}x-3+4x^2+2x-4\\\\=4x^2+\left(\dfrac{1}{4}x+2x\right)+(-3-4)=4x^2+2\dfrac{1}{4}x-7

3 0
3 years ago
Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret.
jek_recluse [69]

Answer:

\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

Step-by-step explanation:

Let X be a set of all students in your class. The set X is the domain. Denote

                                        C(x) -  ' \text{$x $ has a cat}'\\D(x) -  ' \text{$x$ has a dog}'\\F(x) -  ' \text{$x$ has a ferret}'

\mathbf{a)}

Consider the statement '<em>A student in your class has a cat, a dog, and a ferret</em>'. This means that \exists x \in X so that all three statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                         \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)

\mathbf{b)}

Consider the statement '<em>All students in your class have a cat, a dog, or a ferret.' </em>This means that \forall x \in X at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                        \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)

\mathbf{c)}

Consider the statement '<em>Some student in your class has a cat and a ferret, but not a dog.' </em>This means that \exists x \in X so that the statements C(x), F(x) are true and the negation of the statement D(x) . We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                      \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)

\mathbf{d)}

Consider the statement '<em>No student in your class has a cat, a dog, and a ferret..' </em>This means that \forall x \in X none of  the statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as a negation of the statement in the part a), as follows

\neg \left( \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\right) \iff \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)

\mathbf{e)}

Consider the statement '<em> For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.' </em>

This means that for each of the statements C, F and D there is an element from the domain X so that each statement holds true.

We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

           \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

5 0
4 years ago
3(x - 5) = 15<br> Unknown variable
Alex_Xolod [135]

Answer: x = 20/3 = 6.667

Step-by-step explanation:

 Solve  :    3x-20 = 0

Add  20  to both sides of the equation :

                     3x = 20

Divide both sides of the equation by 3:

                    x = 20/3 = 6.667

3 0
3 years ago
Can someone solve this please ;)​
Murljashka [212]

Answer:

see explanation

Step-by-step explanation:

Since the triangles are congruent then corresponding sides and angles are congruent.

In Δ GHJ , GH = GJ thus the triangle is isosceles and the base angles are congruent, thus

∠ J = ∠ H = 67°

∠ T and ∠ H are corresponding and congruent, thus

∠ T = 67°

The sum of the 3 angles in a triangle = 180°, thus

∠ G = 180° - (67 + 67)° = 180° - 134° = 46°

ST and GH are corresponding sides and congruent, thus

ST = 5 cm

HJ and TU are corresponding sides and congruent, thus

HJ = 3.7 cm

Thus

∠ J = 67°, ∠ T = 67°, ∠ G = 46°, ST = 5 cm, HJ = 3.7 cm

6 0
3 years ago
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