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garri49 [273]
3 years ago
5

5(x + 9) + 53 = 14 − 6x

Mathematics
2 answers:
laiz [17]3 years ago
7 0

Answer:

-84/11

Step-by-step explanation:

denis-greek [22]3 years ago
6 0

Answer:

x = -84/11

Step-by-step explanation:

Hope it's right

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Solve the inequality and state the answer in interval notations 3x-7 ≤ 14
emmainna [20.7K]

Answer:

Step-by-step explanation:

3x-7 ≤ 14

3x≤ 14+7

3x≤ 21

x≤ 21/3 = 7

=> x = (-∞;7]

That means that x can be any negative number and any positivi number (1,2,3...7)

4 0
4 years ago
Use the distributive property to remove the parenthese:<br> (-2+5u+v) (-8)
aleksandrvk [35]

Answer:

16 - 40u - 8v

Step-by-step explanation:

(- 2 + 5u + v)(- 8) ← multiply each term in the parenthesis by - 8

= 16 - 40u - 8v

3 0
3 years ago
What are the pros and cons of compasses and straightedges
kotegsom [21]
You're able to recieve more exact measurements and it helps for keeping your work clean and organzied. 
8 0
3 years ago
Technetium-99m is used as a radioactive tracer for certain medical tests. It has a half-life of 1 day. Consider the function T w
Alinara [238K]

Answer:

The expression that represents the number of days until only 10% remains is T((d) 10 %) =100×(\frac{1}{2} )^{3.322}.

Step-by-step explanation:

The equation for half life is of the form

A = A₀×(\frac{1}{2} )^{\frac{t}{h} }.........................................................................(1)

Where

A = Final amount

A₀ = Initial amount

t = Time

h = Half life

For the equation T(d) = 100×2⁽⁻²⁾....................................(2)

We have by comparison with the equation for half life

2 ≡ \frac{t}{h}  and and the equation (2) can be written as

Percentage remaining after 2 half lives is

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{2 }

However if the half life of Technetium-99m is 6 hours then we have for one day

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{2 *2}

Therefore an expression that represents the number of days until only 10% remains is

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{\frac{d}{h}  } = 10 %

(\frac{1}{2} )^{\frac{d}{h}  } =\frac{1}{10}

= ㏑(\frac{1}{2} )^{\frac{d}{h}  }  = ㏑(\frac{1}{10})

= \frac{d}{h}×㏑(\frac{1}{2} ) = ㏑(\frac{1}{10})

\frac{d}{h} = \frac{ln(\frac{1}{10}) }{ln(\frac{1}{2} )} = 3.322

Therefore the expression for the number of days 10 % of Technetium-99m will be remaining is

T((d) 10 %) =100×(\frac{1}{2} )^{3.322}

3 0
4 years ago
Write a situation for 15x-20&lt;130 and solve
Len [333]
15x-20\ \textless \ 130 \\ \\ 15x \ \textless \  130 + 20 \\ \\ 15x \ \textless \  150 \\ \\ x \ \textless \   \frac{150}{15} \\ \\ x \ \textless \  10 \\ \\

Answer: x < 10 
3 0
3 years ago
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