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Anni [7]
3 years ago
9

Find the roots of this equation 7xsquare - 5x - 2 = 0

Mathematics
1 answer:
Sophie [7]3 years ago
7 0

Answer:

x = 1, -2/7

Step-by-step explanation:

You could use the quadratic equation but this can be factored into

(7x + 2) (x - 1).  You can verify that by multiplying it out.

Since (7x + 2) (x - 1) = 0, if either factor is 0 then the equation would be equal to 0, thus we get x = 1, -2/7

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Amy is using a drawing program to complete a construction with which she is almost finished. Which construction is she completin
Vedmedyk [2.9K]

Given options : Two intersecting circles are drawn with a radius in each marked. the image will be linked.

Given options : An equilateral triangle inscribed in a circle

A square inscribed in a circle

A regular pentagon inscribed in a circle

A regular hexagon inscribed in a circle.

<u>Note. When we join an intersection point of two circles and centers of the circles it would form an equilateral triangle that would be inscribe inside a common portion of both circles..</u>

Therefore, an equilateral triangle inscribed in a circle would be correct option.

She is completing an equilateral triangle inscribed in a circle.

4 0
3 years ago
Please help me with number 8. Thank you so much
Nastasia [14]
The are of the trapezoid is 125

4 0
3 years ago
Read 2 more answers
What are the steps to solve for: 1/5(x+5)=1/4(x-3)+1
andrew11 [14]
Step 1. Simplify \frac{1}{5} (x+5) to \frac{x+5}{5}

\frac{x+5}{5} = \frac{1}{4} (x-3)+1

Step 2. Simplify \frac{x+5}{5} to 1+ \frac{x}{5}

1+ \frac{x}{5} = \frac{1}{4} (x-3)+1

Step 3. Simplify \frac{1}{4} (x-3) to \frac{x-3}{4}

1+ \frac{x}{5} = \frac{x-3}{4} +1

Step 4. Cancel 1 on both sides

\frac{x}{5} = \frac{x-3}{4}

Step 5. Multiply both sides by 20 (the LCM of 5,4)

4x=5(x-3)

Step 6. Expand

4x=5x-15

Step 7. Subtract 5x from both sides

4x-5x=-15

Step 8. Simplify 4x-5x to -x

-x=-15

Step 9. Multiply both sides by -1

x=15

Done! :) Hope this helps! :)








3 0
3 years ago
Help me please ASAP please please
Likurg_2 [28]
X > -2....................
5 0
3 years ago
Let f(x)=x+a/x+b such that f(f(1)=0 &amp; f(2)=-3 then a&amp;b resctivily are.
Yuliya22 [10]

Answer:

The value of a = - 1  ,   And b = \frac{ - 7 }{3}    

Step-by-step explanation:

Given as :

Function f(x) = \frac{(x + a)}{(x + b)}

And f(f(1)) = 0     And f(2) = - 3

Now For , x = 2 , y = - 3

I.e  f(2) = \frac{(2 + a)}{(2 + b)}

or,  - 3 = \frac{(2 + a)}{(2 + b)}

I.e 2 + a = - 6 - 3b

Or, a + 3b = - 8               ....... 1

Again  f(f(1)) = 0

So,  \frac{(1 + a)}{(1 + b)} = 0

Or,  1 + a = 0

∴    a = - 1

So , put htis value of a i n eq 1 , we get value of  b

So , - 1 + 3b = - 8

Or,   3b = - 7

∴       b = \frac{ - 7 }{3}

Hence The value of a = - 1  ,   And b = \frac{ - 7 }{3}    Answer

3 0
3 years ago
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