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Deffense [45]
3 years ago
10

What are the solutions to the quadratic equation 5x2 + 60x = 0?

Mathematics
2 answers:
Reptile [31]3 years ago
6 0

Answer:

It's A.

Step-by-step explanation:

5x2 + 60x = 0

5x is the GCF so:

5x(x + 12 ) = 0

5x = 0, x + 12 = 0

x =0, x = -12.

ioda3 years ago
4 0

Answer:

The correct option is 1.

Step-by-step explanation:

The given quadratic equation is

5x^2+60x=0

Taking out the common factors.

5x(x+12)=0

Using zero product property, equate each factor equal to 0.

5x=0\Rightarrow x=0

x+12=0\Rightarrow x=-12

The solutions of the given equations are x=0 and x=-12.

Therefore the correct option is 1.

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1 1/3 x 5/8 <br> Please help, no links or out of context answers, thank you!
natta225 [31]

Answer:

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Step-by-step explanation:

multiply the whole number (1) by the denominator (3) and then add it to the numerator (1).

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5 0
2 years ago
Read 2 more answers
Find the 13th term of the arithmetic sequence -3x – 1,42 + 4,112 + 9, ...
Strike441 [17]

Answer:

The 13th term is 81<em>x</em> + 59.

Step-by-step explanation:

We are given the arithmetic sequence:

\displaystle -3x -1, \, 4x +4, \, 11x  + 9 \dots

And we want to find the 13th term.

Recall that for an arithmetic sequence, each subsequent term only differ by a common difference <em>d</em>. In other words:

\displaystyle \underbrace{-3x - 1}_{x_1} + d = \underbrace{4x + 4} _ {x_2}

Find the common difference by subtracting the first term from the second:

d = (4x+4) - (-3x - 1)

Distribute:

d = (4x + 4) + (3x + 1)

Combine like terms. Hence:

d = 7x + 5

The common difference is (7<em>x</em> + 5).

To find the 13th term, we can write a direct formula. The direct formula for an arithmetic sequence has the form:

\displaystyle x_n = a + d(n-1)

Where <em>a</em> is the initial term and <em>d</em> is the common difference.

The initial term is (-3<em>x</em> - 1) and the common difference is (7<em>x</em> + 5). Hence:

\displaystyle x_n = (-3x - 1) + (7x+5)(n-1)

To find the 13th term, let <em>n</em> = 13. Hence:

\displaystyle x_{13} = (-3x - 1) + (7x + 5)((13)-1)

Simplify:

\displaystyle \begin{aligned}x_{13} &= (-3x-1) + (7x+5)(12) \\ &= (-3x - 1) +(84x + 60) \\ &= 81x + 59 \end{aligned}

The 13th term is 81<em>x</em> + 59.

3 0
3 years ago
Solve for a brainliest in 10 min
horsena [70]

1)x+8=16

x= 16-8

x=8

8 0
2 years ago
1/4 of the seventh grade students went on a field trip to the Alamo. If there a 388 students in the seventh grade, how many stud
Afina-wow [57]

Answer: 97 students

Step-by-step explanation:

Simple! We need to split 388 into four equal parts, so

388 / 4 = 97

Hope this helps!

5 0
3 years ago
The following results come from two independent random samples taken of two populations.
photoshop1234 [79]

Answer:

(a)\ \bar x_1 - \bar x_2 = 2.0

(b)\ CI =(1.0542,2.9458)

(c)\ CI = (0.8730,2.1270)

Step-by-step explanation:

Given

n_1 = 60     n_2 = 35      

\bar x_1 = 13.6    \bar x_2 = 11.6    

\sigma_1 = 2.1     \sigma_2 = 3

Solving (a): Point estimate of difference of mean

This is calculated as: \bar x_1 - \bar x_2

\bar x_1 - \bar x_2 = 13.6 - 11.6

\bar x_1 - \bar x_2 = 2.0

Solving (b): 90% confidence interval

We have:

c = 90\%

c = 0.90

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.90

\alpha = 0.10

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.10/2}

z_{\alpha/2} = z_{0.05}

The z score is:

z_{\alpha/2} = z_{0.05} =1.645

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.645 * \sqrt{0.0735+0.2571}

2.0 \± 1.645 * \sqrt{0.3306}

2.0 \± 0.9458

Split

(2.0 - 0.9458) \to (2.0 + 0.9458)

(1.0542) \to (2.9458)

Hence, the 90% confidence interval is:

CI =(1.0542,2.9458)

Solving (c): 95% confidence interval

We have:

c = 95\%

c = 0.95

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.95

\alpha = 0.05

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.05/2}

z_{\alpha/2} = z_{0.025}

The z score is:

z_{\alpha/2} = z_{0.025} =1.96

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.96 * \sqrt{0.0735+0.2571}

2.0 \± 1.96* \sqrt{0.3306}

2.0 \± 1.1270

Split

(2.0 - 1.1270) \to (2.0 + 1.1270)

(0.8730) \to (2.1270)

Hence, the 95% confidence interval is:

CI = (0.8730,2.1270)

8 0
3 years ago
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