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2 years ago

A quadratic equation has exactly one real number solution.Which is the value of its discriminant ?-1,0,1 or 2

Mathematics

1 answer:

Ivanshal [37]2 years ago

4 0

Answer:

The value of the discriminant is 0

Step-by-step explanation:

The discriminant of a quadratic equation is calculated as thus:

b² - 4ac and it is represented by letter D.

Hence, D = b² - 4ac..

This discriminant can take either of 3 range of values.

1. D > 0. This implies that the discriminant has a value greater than 0 and this means that the root has 2 different real solutions

2. D = 0. This implies that the discriminant equals 0 and this means that the root has only 1 solution.

3. D < 0. This implies that the discriminant is less than 0 and this means that the root has 2 different complex solutions.

Hence, D = 0 is the correct answer to this question.

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Round 5.492 to the place value of the underlined digit.

mars1129 [50]

If it has a line under the 9, then it needs to be 5.49, but if it has a line under the 4 it needs to be rounded up to 5.5 since the number behind it is above 5.

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2 years ago

0.32 with the 2 repeating as a fraction

pashok25 [27]

Answer:

29/90

Step-by-step explanation:

Use the formula:

((DN x F) - NRP)/D

DN = Decimal Number

F = 10 if one repeating number, 100 if two repeating numbers, 1000 if three repeating numbers, etc.

NRP = Non-repeating part of decimal number.

D = 9 if one repeating number, 99 if two repeating numbers, 999 if three repeating numbers, etc.

((0.32 x 10) - 0.3)/9

= 2.9/9 = 29/90

3 0

1 year ago

What is the simplified version of 3\sqrt{135}

Aleonysh [2.5K]

Answer:

The simplified version of $\sqrt[3]{135}$ is $3\sqrt[3]{5}$ .

Step-by-step explanation:

The given expression is

$\sqrt[3]{135}$

According to the property of radical expression.

$\sqrt[n]{x}=(x)^{\frac{1}{n}}$

Using this property we get

$\sqrt[3]{135}=(135)^{\frac{1}{3}}$

$\sqrt[3]{135}=(27\times 5)^{\frac{1}{3}}$

$\sqrt[3]{135}=(3^3\times 5)^{\frac{1}{3}}$

$\sqrt[3]{135}=(3^3)^{\frac{1}{3}}\times (5)^{\frac{1}{3}}$ $[\because (ab)^x=a^xb^x]$

$\sqrt[3]{135}=3\times \sqrt[3]{5}$ $[\because \sqrt[n]{x}=(x)^{\frac{1}{n}}]$

$\sqrt[3]{135}=3\sqrt[3]{5}$

Therefore the simplified version of $\sqrt[3]{135}$ is $3\sqrt[3]{5}$ .

7 0

2 years ago

Read 2 more answers

-2(x-3) >-8 What is the answer

Vlad [161]

Answer:

x < 7

Step-by-step explanation:

First apply the -2 onto each term that it is being multiplied onto.

-2x + 6 > - 8

Now subtract six from both sides.

-2x > -14

Now divide by -2.

x < 7

REMEMBER

When doing greater than and less than equations like this, if you divide by a negative you must flip the greater than/less than.

5 0

2 years ago

Given the height, h, and the volume v of a certain cylinder, alex uses the formula v ph r 5 !§ to compute its radius, r, to be 1

dmitriy555 [2]

We know that
volume of a cylinder=pi*r²*h
r=√[V/(pi*h)]
r=10 m

A second cylinder has
V2=V-----> volume of the second cylinder
h2=25*h----> height of the second cylinder
r2----> radius of the second cylinder
r2=√[V2/(pi*h2)]----> r2=√[V/(pi*25*h)]---> r2=(1/5)*√[V/(pi*h)

r2=(1/5)*10-----> r2=2 m

the answer is
the radius of the second cylinder is 2 m

7 0

2 years ago