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3 years ago

Is 5 times 5/6 greater than 5

Mathematics

1 answer:

jenyasd209 [6]3 years ago

3 0

Answer:

Step-by-step explanation:

since 5/6 is not a whole it should be a little less than 5

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Find all zeros and write in factored form (x-5)(3x^2-8x+6) =0

Snezhnost [94]

Answer:

x = 5, 4/3+i√2/3, 4/3-i√2/3.

Step-by-step explanation:

(x-5)(3x^2-8x+6) =0

This is in factored form already. 3x^2 - 8x + 6 cannot be factored, it is prime.

One zero is x = 5.

3x^2-8x+6 = 0 Using the quadratic formula:

x = [8 +/- sqrt (64 - 4*3*6]) / 6

x = 4/3 +/- sqrt (-8) / 6

x = 4/3 +/- 2sqrt2 i / 6

x = 4/3 +/- i√2 / 3.

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3 years ago

|2x-1|=5 please help

VikaD [51]

Answer:

X=3

Step-by-step explanation:

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3 years ago

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PLZ PLZ PLZ PLZ HELP - The legs of a right triangle are 3 units and 6 units. What is the length of the hypotenuse? Round your an

Oduvanchick [21]

Answer:

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6 0

3 years ago

Please help the following questions

34kurt

Answer:

Question 7. Option B,

Question 8. Option C

Step-by-step explanation:

Question 7.

Let us create a proportionality, and solve for the value of s, through cross multiplication;

$3 / 12 = s / 15,\\3 * 15 = 12 * s,\\12 * s = 45,\\\\s = 3.75\\\\Conclusion - Option B$

Question 8.

We can solve this problem in a similar manner, through setting a proportionality and solving for the length of the shadow of the tree, which we can pose as x;

$5 / 6 = 25 / x,\\5 * x = 6 * 25,\\5x = 150,\\\\x = 30 ft\\\\Conclusion - Option C$

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3 years ago

I need help on this pleasssee!!

aev [14]

Answer:

Option D. $x=-3$

Step-by-step explanation:

we have

$f(x)=\frac{5x^{4}-7}{x+3}$

we know that

The denominator cannot be equal to zero

therefore

To find the equation of the vertical asymptote equate the denominator to zero

$(x+3)=0$

$x=-3$ -----> equation of the vertical asymptote

8 0

4 years ago