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omeli [17]
3 years ago
14

The answer should be a fraction

Mathematics
2 answers:
schepotkina [342]3 years ago
6 0
<span>Hello! : )

The correct answer to your question is...</span> 2^{3} y^{5}<span>

Evidence:
Step 1:     Simplify </span>y^{2}<span>/5</span><span>

( 40 multiplied by</span> y^{2}/5 multiplied by y^{3}<span> )
</span>

Step 2:     y^{2}<span> multiplied by </span>^{3}<span> = y(2 + 3) = </span>y^{5}


So the correct answer is... 2^{3}  y^{5}

I hope this helped! : )
Please Rate & Thank!
Please mark as Brainliest!

Have a wonderful day! : )
Vedmedyk [2.9K]3 years ago
5 0
Ok so 40y2/50y3.... you are going to cancel out the common factor (10)  
 
4y2/5y3..
now apply the exponent rule, which is : xa /xb = 1/ xb-a

so...y2/y3 = 1/ y3 - 2 = 1/y

ANSWER : 4/5y
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So what they are telling you is that 5 is three higher than the variable n.
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so n is 2. 
4 0
3 years ago
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ONLY answer if yk it i’m giving best answer brainliest!!
S_A_V [24]

Answer:

-x^4-10x+1/10

Step-by-step explanation:

standard form goes from biggest power to smallest power

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please help me and explain. what is the value of <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7By%7D%20" id="TexFormula1"
rosijanka [135]
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multiplying both sides by x^2
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simplifying
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8 0
3 years ago
The coordinate plane below represents a city. Points A through F are schools in the city.
Usimov [2.4K]
Part A. The technique on how to find the equation that only applies to point D and E, is to create a line or curve that only includes two of these points. In this case, I created a random parabola that isolates points C and F from the rest of the points. First, we have to find the equation of the parabola through its general forms:

(x - h)² = +/-4a(y-k)      or    (y - k)² = +/-4a(x - h)

For parabolas drawn like that in the picture, the general form is (x - h)² = +4a(y-k), where the vertex is (h,k) and a is the distance from the vertex to the focus. From the picture, the vertex is at (0,3). Then, we use point D(-2,4) to determine a:

(-2 - 0)² = +4a(4 - 3)
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So, the equation of the parabola is:

x² = 4(y - 3)
x² = 4y - 12

Part B. Point D was already verified above. Now for point E(2,4)
x² = 4y - 12
2² ? 4(4) - 12
4 ? 4
4 = 4

Part C. For y < 7x − 4, ignore the equality symbol first and graph the line. Assign random values of x, then you get corresponding values of y. Plot them as shown in the second picture. The line is shown in red. Next, test the equation by choosing a random point. Let's choose the purple point at (4,3).

3 ? 7(4) − 4
3 ? 24
3 < 24

Thus, it applies to the purple point, and all the other areas to that area. The shaded region are all solutions of the inequality. So, Erica is only interested in points E, C and F. 

5 0
3 years ago
Can someone help me with this and show their work and how they found the answer?
Semenov [28]

Answer:

2kg

Step-by-step explanation:

Let the sum of the other 2 bricks be R. Given that the heaviest brick weighs 2/3 times as much as the other 2 bricks in total, then the weight of the heaviest brick H in terms of the weight of the other 2 will be

H = 2/3 * R

= 2R/3

Given that the 3 bricks weigh 5kg in total then,

2R/3 + R = 5

Multiply through the equation by 3

2R + 3R = 15

5R = 15

Divide both sides by 5

R = 15/5

= 3

The weight of the heaviest brick is 2R/3. Since R = 3, the heaviest block weighs

= 2 * 3/3

= 2kg

4 0
3 years ago
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