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vazorg [7]
3 years ago
13

3

formula">+7.1=2.5b+11.5
Mathematics
1 answer:
vlada-n [284]3 years ago
8 0
( \beta +7.1=2.5b+11.5)*10 \\ 10b+71=25b+115 \\ 10b-25b=115-71 \\ -15b=44 \\ b= \frac{-44}{15}
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Find the two correct values of x in the equation 3x² = y when y = 75.​
valentina_108 [34]

Answer:

5 and -5

Step-by-step explanation:

3x² = 75

x² = 75/3

x² = √25

x = 5 and - 5

3 0
2 years ago
Which equation is not equivalent to the other two
Liula [17]

Answer:

y=-4-3x

Step-by-step explanation:

y = -3x - 4

y = -3x + (-4)

y = -4 + 3x

y = -4 - 3x

7 0
3 years ago
PLEASE HELP?! IM SO CONFUSED!!
Andre45 [30]

Answer:

C.

Step-by-step explanation:

-3 7/8 rounds down to -4

4.63 rounds up to 5

-4 x 6 + 5

-24 + 5

-19

her answer (-19.14) also rounds to -19

So yes, it is reasonable and the correct response is C.

5 0
2 years ago
The congruence theorem that can be used to prove ALON
Arisa [49]

Answer:

SSS is the congruence theorem that can be used to prove Δ LON  is congruent to Δ LMN ⇒ 1st answer

Step-by-step explanation:

Let us revise the cases of congruence

  • SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
  • SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
  • ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
  • AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
  • HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ  

In triangles LON and LMN

∵ LO ≅ LM ⇒ given

∵ NO ≅ NM ⇒ given

∵ LN is a common side in the two triangles

- That means the 3 sides of Δ LON are congruent to the 3 sides

   of Δ LMN

∴ Δ LON ≅ LMN ⇒ by using SSS theorem of congruence

SSS is the congruence theorem that can be used to prove Δ LON  is congruent to Δ LMN

3 0
3 years ago
Find the positive value of x in the geometric sequence 7x-2,4x+4,3x
Goryan [66]
Hello,


Let's assume k the ratio.


 \left \{ {{3x=k*(4x+4)} \atop {4x+4=k*(7x-2)}} \right. \\\\
 \left \{ {{3x-4kx=4k} \atop {4x-7kx=-2k-4}} \right. \\\\
 \left \{ {{x(3-4k)=4k} \atop {x(4-7k)=-2k-4}} \right. \\\\
 \left \{ {{x=\dfrac{4k}{3-4k} \atop {x=\dfrac{-2k-4}{4-7k} \right. \\\\

\dfrac{4k}{3-4k}=\dfrac{-2k-4}{4-7k}\\\\

36k^2-6k-12=0\\
6k^2-k-2=0\\

\boxed{k= \dfrac{2}{3} \ or\ k= -\dfrac{1}{2}}\\\\

x=\dfrac{4*\dfrac{2}{3} }{3-4*\dfrac{2}{3} }=8\\
x=\dfrac{4*\dfrac{-1}{2} }{3-4*\dfrac{-1}{2} }=-\dfrac{2}{5}\\







\boxed{x= 8 \ or\ x= -\dfrac{2}{5}}\\\\





5 0
3 years ago
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