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3 years ago

Solve 2(x - 5) + 11 = 17

Mathematics

1 answer:

mrs_skeptik [129]3 years ago

5 0

Answer:

x=8 and is not extraneous

Step-by-step explanation:

2(x-5)+11=17

2x-10+11=17

2x+1=17

2x=17-1

2x=16

x=8

extraneous means it is a solution, but it is not possible in the equation, to determine if it is extraneous, plug it in to the original equation

2(8-5)+11=17

2(3)+11=17

6+11=17

17=17

since it works in the orignial equation it is not extraneous

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Find the value of 5x-6 given that 2x-7=9

erik [133]

Answer:

Step-by-step explanation:

Given that

2x - 7 = 9 ← solve for x

Add 7 to both sides

2x = 16 ( divide both sides by 2 )

x = 8

Substitute x = 8 into 5x - 6

5x - 6 = (5 × 8) - 6 = 40 - 6 = 34

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3 years ago

Jennifer is organising a street party. she needs to make 300 cakes. the recipe consist of : 50g icing, 150g butter, 120g caster

joja [24]

Answer:

1.25 kg of icing sugar

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Step-by-step explanation:

300/12 x (weight of ingredient)

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3 years ago

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What is the answer when you simplify 44/80

irina1246 [14]

Answer:

11/20 is simplified 44/80

Step-by-step explanation:

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3 0

3 years ago

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Select True or False for each statement.

SSSSS [86.1K]

$\left( \dfrac 1 {64} \right)^{- 5/6} =64^{5/6} = (\sqrt[6]{64})^5 = 2^5 =32$

TRUE

$\sqrt[5]{36^4}=36^{4/5}$

which surely isn't 36. FALSE

$\sqrt{12} - \dfrac 2 5 \sqrt{75} = 2 \sqrt{3} -\dfrac 2 5 (5) \sqrt{3} = 0$

FALSE

For the fourth one we have a

$\sqrt{98b} + \sqrt{2b}$

which isnt

$10\sqrt{b}$

so this is FALSE.

$\dfrac{1}{(\sqrt 5 - \sqrt 6)^2}$

$= \dfrac{1}{(\sqrt 5 - \sqrt 6)^2} \cdot \dfrac{(\sqrt 5 + \sqrt 6)^2}{(\sqrt 5 + \sqrt 6)^2}$

$= \dfrac{(\sqrt 5 + \sqrt 6)^2}{ ( (\sqrt 5 - \sqrt 6)(\sqrt 5 + \sqrt 6))^2}$

$= \dfrac{(\sqrt 5 + \sqrt 6)^2}{( 5-6)^2}$

$=(\sqrt 5 + \sqrt 6)^2$

No fractions in that one so FALSE.

3 0

3 years ago

Rewrite the equation 10=1,000 with an exponet that makes it true .

inn [45]

10^3= 1,000
hope this helps :)

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3 years ago

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