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eduard
2 years ago
10

Can any equation in point-slope form also be written in slope-intercept form?

Mathematics
1 answer:
Gekata [30.6K]2 years ago
3 0

Answer:

Yes

Step-by-step explanation:

The way the equation is written in point-slope form is:

y-y1=m(x-x1)

The way we write an equation in slope-intercept form is:

y=mx+b

We can change point-slope form to slope-intercept form by first distributing the m and then adding y1 to both sides of the equation. When we do this we will get:

y=mx-(m*x+y1) where (m*x+y1) = b

which is now in slope intercept form.

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Factor completely. x 3 + 6x 2 - 4x - 24
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You can factor (x^3 and 6x^2) and (-4x and -24).

So x^2(x+6) - 4(x+6)

(x^2-4)(x+6)

(x-2)(x+2)(x+6)
5 0
3 years ago
What is the solution of 5 x + 1] + 6 = 21?
Mice21 [21]

Answer:

x = -4 or x = 2

Step-by-step explanation:

5 0
3 years ago
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An ulcer patient has been told to avoid acidic foods. If he drinks coffee, with a pH of 5.0, it bothers him, but he can tolerate
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The equation should look like :
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3 0
3 years ago
Minstrel Manufacturing uses a job order costing system. During one month, Minstrel purchased $198,000 of raw materials on credit
Diano4ka-milaya [45]

Answer:

$440,000

Step-by-step explanation:

Direct material:

= $195,000 - $30,000

= $165,000

Direct labor:

= $150,000 - $40,000

= $110,000

Manufacturing overhead:

150% of direct labor cost.

= $110,000 x 150 ÷ 100

= $16,500,000 ÷ 100

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The total manufacturing costs added during the period is: <u>$440,000</u>

5 0
3 years ago
Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 colleg
Fofino [41]

Answer:

The probability that 75% or more of the women in the sample have been on a diet is 0.037.

Step-by-step explanation:

Let <em>X</em> = number of college women on a diet.

The probability of a woman being on diet is, P (X) = <em>p</em> = 0.70.

The sample of women selected is, <em>n</em> = 267.

The random variable thus follows a Binomial distribution with parameters <em>n</em> = 267 and <em>p</em> = 0.70.

As the sample size is large (n > 30), according to the Central limit theorem the sampling distribution of sample proportions (\hat p) follows a Normal distribution.  

The mean of this distribution is:

\mu_{\hat p} = p = 0.70

The standard deviation of this distribution is: \sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028

Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

P(\hat p \geq 0.75)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}} \geq \frac{0.75-0.70}{0.028}) =P(Z\geq0.179)=1-P(Z

**Use the <em>z</em>-table for the probability.

P(\hat p \geq 0.75)=1-P(Z

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.

7 0
3 years ago
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