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3 years ago

The exterior angle of a triangle is 164 degree and one

Mathematics

1 answer:

nataly862011 [7]3 years ago

5 0

Answer:

• Let the other interior angle be x

${ \rm{x + 98 \degree = 164\degree}} \\ \\ { \rm{x = 164\degree - 98\degree}} \\ \\ { \boxed{ \rm{x = 66\degree}}}$

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What are the possible zeros for f(x) equals X to the third power minus 2X to the second power minus 4X +15

Volgvan

The roots of the polynomial are done by using a calculator will be -2.3671, 2.1835 + 1.2527i, and 2.1835 – 1.2527i.

<h3>What are the roots of the polynomial?</h3>

The polynomial is given below.

$\rm f(x) = a_0x^n + a_1 x^{n - 1}+ ....+ a_{n-2}x^2 + a_{n-1}x + a_nx^0$

The roots of the polynomial are equal to n or less than n.

And if one imaginary root is a + bi, then the other imaginary root will be a – bi.

The roots of the polynomial are done by using a calculator.

x = -2.3671, 2.1835 + 1.2527i, 2.1835 – 1.2527i

More about the roots of the polynomial link is given below.

brainly.com/question/1514617

#SPJ1

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2 years ago

Who can answer my question in square LMNO find the length of LM

AlexFokin [52]

Answer:

10 units

Step-by-step explanation:

LMNO is a square.

All sides of square are equal

MN = LM

3x + 1 = x + 7

3x - x = 7 - 1

2x = 6

x = 6/2

x = 3

LM = x + 7 = 3 + 7 = 10

LM = 10 units

8 0

3 years ago

What is the range of f(x)?

slavikrds [6]

It -2,-1 I hope this help

6 0

3 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

4 years ago

Identify the slope of the line for the equation y = −6x − 9.

Pepsi [2]

Answer:

-6

Step-by-step explanation:

This line is written in <em>slope intercept</em> formula! (y=mx+b)

The variable <u>m</u> in <em>slope intercept</em> formula is the <u>slope</u>, and you have been given the value of -6 as slope!

4 0

3 years ago

Read 2 more answers