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Blababa [14]
3 years ago
11

What is the value of x a. 8.125 b .7.25c.6.125 d.3.25

Mathematics
2 answers:
MariettaO [177]3 years ago
7 0

Answer:


Step-by-step explanation:


ahrayia [7]3 years ago
3 0
What's the equation with x? I can't solve it without knowing where to find x....
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Sonbull [250]

Answer:

i do not think it is possible to delete a question after you have posted it. if i were you i probably would have just edited it and save myself the emmbarasment

Step-by-step explanation:

8 0
2 years ago
Anyone know how to do this?
IrinaVladis [17]
H(x) = x - 5

However, it had a domain of all real numbers such that x does not equal 3, since that would cause you to divide by 0. 
7 0
2 years ago
Read 2 more answers
Which values of <br> P<br> and <br> Q result in an equation with no solutions?<br> 83x+P=83x+Q
aleksandr82 [10.1K]

83x+P=83x+Q

subtract 83x from each side

P=Q

if P does not Q there are no solutions

the lines would be parallel and never intersect if Q does not equal P

6 0
2 years ago
Side lengths and angle measures in similar figures ​
Ksivusya [100]

Step-by-step explanation:

You can see the full detailed explanation in the attached files

7 0
3 years ago
the sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger
Reil [10]
<h2>Answer:</h2>

x\geq 3 \ and \ x+1 \geq 4

<h2>Step-by-step explanation:</h2>

The question in this problem is:

<em>The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?</em>

<em />

First of all, let's name the first variable x which is the smaller number. Accordingly, the lager number will be x+1 given that those numbers are consecutive. On the other hand<em> at most </em>conveys the idea of an inequality, which is:

\leq \\ which \ means \ less \ than

So:

1. The sum of 2 consecutive integers can be written as:

v+(v+1)

2. Nine times the smaller and 5 times the larger can be written as:

9v-5(v+1)

Finally, the whole statement:

The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger:

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\ \frac{6}{2} \leq \frac{2x}{2} \\ \\ 3 \leq x \\ \\ x\geq 3 \\ \\ and \\ \\ x+1 \geq 4

The two numbers are:

x\geq 3 \ and \ x+1 \geq 4

6 0
3 years ago
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