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3 years ago

Please do for me thank you I am

Mathematics

1 answer:

Licemer1 [7]3 years ago

8 0

4.) top row: 1, 2, 4, 6, 8, 10

bottom row: 25, 50, 100, 150, 200, 250

5.) top row: 1, 2, 4, 6, 8, 10

bottom row: 50, 100, 200, 300, 400, 500

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Kelly simplified this power of a product.

Natalija [7]

Answer:

the answer is D

Step-by-step explanation:

i got it right

5 0

3 years ago

Read 2 more answers

5. Find the discriminant of 15x2 = 4x – 1 and describe

diamong [38]

Answer:

a) The discriminant of the equation = - 44

b)The nature of the roots will be imaginary.

c) $x = \frac{2 +\sqrt{11} i}{15} or, x = \frac{2 - \sqrt{11} i}{15}$

Step-by-step explanation:

Here, the given expression is $15x^{2} = 4x -1$

or, $15x^{2} - 4x + 1 = 0$

Now the discriminant (D) of a quadratic equation $ax^{2} +b x + c = 0$

D = $b^{2} - 4ac = (-4)^{2} - 4(15) (1) = 16 - (60) = -44$

Hence, the discriminant of the equation = - 44

As D< 0, so the roots will be imaginary.

Now,by quadratic formula : $x = \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}$

So, here $x = \frac{-(-4) \pm \sqrt{D} }{2a} = \frac{4 \pm \sqrt{(-44 )} }{30}$

So, either $x = \frac{4 + \sqrt{(-44 )} }{30} or, x = \frac{4 - \sqrt{(-44 )} }{30}$

or, $x = \frac{2 +\sqrt{11} i}{15} or, x = \frac{2 - \sqrt{11} i}{15}$

8 0

3 years ago

3 is 12% of what number

AleksAgata [21]

3/12% = 25

<em>Therefore, 3 is 12% of 25.</em>

3 0

4 years ago

Read 2 more answers

Water flows from a faucet at a constant rate. Assume that 6 gallons of water are already in a tub by the time we notice the fauc

Thepotemich [5.8K]

The equation would be y=3.2x+6

6 0

3 years ago

Isabel wrote this mystery problem. The quotient is multiple of 6. The dividend is multiple of 9. The divisor is a factor of 12.

Marina86 [1]

Answer:

Consider the expression $q=\frac{a}{b}$ . q is called the quotient, a is is the dividend and b is the divisor.

Since q is a multiple of 6, then q has the form $q=6k$ for some integer k.

Since a is a multiple of 9, then a has the form $a=9s$ for some integer s.

Since b is a factor of 12, then if 12 can be expressed of the form $12=c*b$ , for c an integer. Then b has the form $b=\frac{12}{m}$

Replacing the preview expression in the initial expression we obtain:

$6k=\frac{9s}{\frac{12}{m}}=\frac{9sm}{12}\\6*12k=9sm\\72k=9sm\\72k-9sm=0$

Then $72k-9sm=0,\; for \; k,s,m\in \mathbb{Z}$ is a equation to Isabel's problem.

3 0

3 years ago