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Yuki888 [10]
3 years ago
15

What is the value of x when (6x+1)+(2x+10)=79

Mathematics
1 answer:
Setler [38]3 years ago
5 0

Answer:

x=8.5

Step-by-step explanation:

(6x+1)+(2x+10)=79

6x+2x    1+10

8x +11=79

-11        -11

8x=68

÷8   ÷8

x=8.5

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Rita earns $10 per hour. She puts 5% of her earnings in savings. Write inequality to find how many hours Rita must work to save
Rina8888 [55]

Answer:

Rita must work 50 days  to save at least $25.

Step-by-step explanation:

The amount Rita earns per hour  = $10

The saving percentage =  5%

So, the savings of Rita per hour =  5% of $10

⇒5% of $10  = \frac{5}{100}  \times 10  = 0.5

or Rita saves $0.5 per hour.

The amount she wants to save at minimum  = $25

So, let she works at the minimum of k Days.

⇒\textrm{The number of hours she has to work}  = \frac{\textrm{Amount she needs saving}}{\textrm{Amount saved in 1 hour}}

or, k = \frac{25}{0.5}  = 50

The number of hours she has to work  =  50 days

Hence,  Rita must work 50 days  to save at least $25.

3 0
3 years ago
A line passes through the points (1, 3) and (3, -1) on a coordinate plane. What is the equation of this line in slope-intercept
FromTheMoon [43]

Answer:

y=[-2]x[+][5]

Step-by-step explanation:

\frac{-1-3}{3-1}=\frac{-4}{2} =-2 m= -2

Using the point (1,3) Find the value of b

3= -2(1)+b

3= -2+b

+2  +2

---------

5=b

y= -2x+5

4 0
3 years ago
Read 2 more answers
Find the point q that is 1/4 the distance from a to b
Scorpion4ik [409]
Bro sorry it’s making me answer questions
5 0
3 years ago
Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction
Oliga [24]

The base case is the claim that

\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}

which reduces to

\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86

which is true.

Assume that the inequality holds for <em>n</em> = <em>k </em>; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}

By the induction hypothesis,

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}

Now compare this to the upper bound we seek:

\dfrac{2k+1}{k+1}  > \dfrac{2k+2}{k+2}

because

(2k+1)(k+2) > (2k+2)(k+1)

in turn because

2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0

6 0
2 years ago
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Pls help will give brainliest and 5star pls help really need it
Temka [501]
It would be B or 1,625 because percentages are going to always involve multiplication. if I'm wanting to find a percentage you do .25 times 6,500 which leaves you with 1,625.
8 0
3 years ago
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