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konstantin123 [22]
3 years ago
15

Identify the number of solutions: 5x + 4 = 5x + 4

Mathematics
1 answer:
Murrr4er [49]3 years ago
6 0

Answer:

0,4) : not a solution

(-2,4) ; solution

(0,5) : solution

(-2,7) : solution

(-4,1) : not a solution

(-1,1) : not a solution

(-1.5,3.5) : solution

any point that falls in the grey area is a solution....if ur points lie on a dashed line, it is not a solution...but if ur points lie on a solid line, it is a solution

Step-by-step explanation:

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i think the answer is -33

Step-by-step explanation:

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The graph shows the function f(x)=3^x
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The answer is D.
You have to make the whole numbers equal to then solve for the exponents:
F(x) means y, therefore, y is 9.
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Prove that if xy > 0, then either x > 0 and y > 0 or x < 0 and y < 0
Rufina [12.5K]

The rule of product states that:

  • (+) \cdot (+) = (+)
  • (+) \cdot (-) = (-)
  • (-) \cdot (+) = (-)
  • (-) \cdot (-) = (+)

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Brut [27]

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Step-by-step explanation:

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Read 2 more answers
Please help me. I need help please.
Mice21 [21]

Answer:

The correct options are;

EFGH has 4 congruent sides

Diagonal FH bisects angles EFG and EHG

Angle FEH is congruent to angle FGH

Step-by-step explanation:

1) Given that for a reflection, we have;

The distance of the reflected preimage from the line of reflection = The distance of the reflected image from the line of reflection

Therefore;

The distance of the point E from the line HF = The distance of the point G from the line HF

Also the reflection of an preimage (x, y) about the x-axis, gives an image (x, -y)

We can show that from the length of a line given by the equationl = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}, that the length EH ≅ GH and EF ≅ GF

Therefore since we are given that EH = EF, we have;

EH = GH = GF = EF by the definition of congruency, which gives 4 congruent sides

2) Given that EH = GH = GF = EF and HF = FH by reflective property, we have;

ΔEHF ≅ ΔGHF

∴ ∠GHF ≅ ∠EHF by Congruent Parts of Congruent Triangles are Congruent

Similarly, ∠GFH ≅ ∠EFH

Therefore, ∠GFH = ∠EFH and ∠GHF = ∠EHF

Therefore, diagonal FH bisects angles EFG and EHG

3) Given that ΔEHF ≅ ΔGHF, we have;

Angle FEH is congruent to angle FGH, by Congruent Parts of Congruent Triangles are Congruent

7 0
3 years ago
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