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3 years ago

Please help 30 points!!'

Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

7 0

Answer:

The equation of line perpendicular to given line passing through (-15,-4) is:

$y = -\frac{4}{3}x-24$

Step-by-step explanation:

Given equation is:

$3x-4y = 9$

The given equation is in standard form. We have to convert it into slope intercept form first.

The slope intercept form is:

$y = mx+b$

So,

$3x-4y=9\\3x-9 = 4y\\4y = 3x-9\\\frac{4y}{4} = \frac{3x-9}{4}\\y = \frac{3}{4}x - \frac{9}{4}$

The co-efficient of x is the slope of the line.

m = 3/4

The product of slopes of two perpendicular lines is: -1

Let m1 be the slope of line perpendicular to given line then

$m.m_1 =-1\\\frac{3}{4} . m_1 = -1\\m_1 = -1 * \frac{4}{3} \\m_1 = -\frac{4}{3}$

Putting in slope-intercept form

$y = m_1x+b\\y = -\frac{4}{3}x+b$

To find the value of b, putting (-15,-4) in equation

$-4 = -\frac{4}{3}(-15) +b\\-4 = 20+b\\-4-20 = b\\b = -24$

The equation will be:

$y = -\frac{4}{3}x-24$

Hence,

The equation of line perpendicular to given line passing through (-15,-4) is:

$y = -\frac{4}{3}x-24$

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I only need help with number 9 plz

Varvara68 [4.7K]

Answer:

<h2>Letter C</h2>

Step-by-step explanation:

If we look on the number line we see that there is a point on -3, another point on -1, 2, and 7.

These points tell us our answer, but the question states it doesn't have to be in any order.

As we look at answer A we see -3, 1, 2, and -7. That's incorrect since there is no positive 1 nor an -7.

As we look at answer B we see -3, 7, 1, and -2. That's incorrect since there is no positive 1, and not a -2.

As we look at answer D we see -2, 3, 1, and 7. That's incorrect since there is no positive 2, no positive 3, and no negative 1.

C has to be the correct answer since it's the only answer left and has the 4 numbers that are on the number line.

Therefore, the correct answer is C.

Best of Luck to you.

If you have any questions, feel free to comment below.

6 0

3 years ago

An electrical engineer wishes to compare the mean lifetimes of two types of transistors in an application involving high-tempera

Olin [163]

Answer:

(115.2642, 222.7358).

Step-by-step explanation:

Given data:

type A: n_1=60, xbar_1=1827, s_1=168

type B: n_2=180, xbar_2=1658, s_2=225

n_1 = sample size 1, n_2= sample size 2

xbar_1, xbar_2 are mean life of sample 1 and 2 respectively. Similarly, s_1 and s_2 are standard deviation of 1,2.

a=0.05, |Z(0.025)|=1.96 (from the standard normal table)

So 95% CI is

(xbar_1 -xbar_2) ± Z×√[s1^2/n1 + s2^2/n2]

=(1827-1658) ± 1.96×sqrt(168^2/60 + 225^2/180)

= (115.2642, 222.7358).

8 0

3 years ago

Please try to give me the answers

gladu [14]

Answer:

a = 1, b = 2, c = 3

Step-by-step explanation:

$a + \frac{1}{b + \frac{1}{c} } = \frac{10}{7} \\ \\ a + \frac{1}{b + \frac{1}{c} } = 1 + \frac{3}{7} \\ \\ a + \frac{1}{b + \frac{1}{c} } = 1 + \frac{1}{\frac{7}{3} } \\ \\ a + \frac{1}{b + \frac{1}{c} } = 1 + \frac{1}{2 + \frac{1}{3} } \\ \\ equating \: like \: terms \: on \: both \: sides \\ \\ a = 1 \\ \\ b = 2 \\ \\ c = 3$