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Goryan [66]
3 years ago
5

Complete the following equation using <, >, or = 120% ___ 1

Mathematics
1 answer:
kvasek [131]3 years ago
3 0
To solve these types of problems, we need to convert the two values into the same form.  

We know that 1 = 1.00, as adding zeroes to the end of a decimal doesn't change the value.

To change a decimal into a percentage, we multiply the decimal by 100 and add a percent sign.  

1.00 * 100 = 100%

Therefore, 1 = 100%

Because 120% is greater than 100%, thus, 

120% > 1
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2x-5=x+25 what is the answer for x because am stuck
Stels [109]

2x - 5 = x + 25

<u>-x      </u>   <u>-x         </u>

x - 5  =       25

<u>    +5</u>     <u>      +5 </u>

x       =       30

Answer: x = 30

8 0
3 years ago
Which expression is equivalent to -2<br>,<br>글<br>금<br>1 0 23<br>​
den301095 [7]

\bf \cfrac{-3}{4}\cdot \cfrac{7}{-2}\div\cfrac{3}{-8}\implies \cfrac{-3}{4}\cdot \cfrac{7}{-2}\cdot \cfrac{-8}{3}\implies \cfrac{-3}{4}\cdot \cfrac{-8}{3}\cdot \cfrac{7}{-2} \\\\\\ \cfrac{-3}{3}\cdot \cfrac{-8}{4}\cdot \cfrac{7}{-2}\implies -1\cdot \cfrac{-2}{1}\cdot \cfrac{7}{-2}\implies \cfrac{2}{1}\cdot \cfrac{7}{-2}

3 0
3 years ago
The graph below shows the solution for the following system.
Rudiy27

Answer:

TRUE:

When x≈2.7, the graphs of f(x) and g(x) intersect

f(x)=g(x) when x=0

Step-by-step explanation:

The graphs of two function y=f(x) and y=g(x) are shown in attached diagram.

These two graphs intersect at two points (0,-3) and about (2.7,2.3). This means that

f(0)=g(0)=-3

and

f(2.7)=g(2.7)=2.3

So, x=0, y=-3 is the solution to the system (the solution to the system is ordered pair (x,y), not only x)

Points (1.5,0) and (2,0) are not solutions, because they are not points of graphs intersection.

When x≈2.7, the graphs of f(x) and g(x) intersect (TRUE)

f(x)=g(x) when x=0 (TRUE)

8 0
3 years ago
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Step-by-step explanation:

stat edit sort a calculator

4 0
3 years ago
Prove that the equation x^2+px-1=0 for every p has two different solutions
Sladkaya [172]

Answer:

To prove: The equation x2+px−1=0 has real and distinct roots for all real values of p.

Consider x2+px−1=0

Discriminant D=p2−4(1)(−1)=p2+4

We know p2≥0 for all values of p

⇒p2+4≥0 (since 4>0)

Therefore D≥0

Hence the equation x2+px−1=0 has real and distinct roots for all real values of p

8 0
3 years ago
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