All categories

3 years ago

What is the value of x? Enter your answer in the box. x =

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

8 0

Here you have an equilateral triangle. Because of this you know that all angles are also the same as each other. We know that the angles in an equilateral triangle are 60°.

so let's set 7x + 4 = 60
Subtract 4 from both sides: 7x = 56
Divide both sides by 7: x = 8

You might be interested in

Find the slope of the line. Write your answer in simplest form.

Masja [62]

Answer:

1/6

Step-by-step explanation:

The rise is 1 and the run is 6

7 0

3 years ago

Does anyone know the questions for the dba of pre algebra 2.00?

Oksanka [162]

Answer:

If this is FLVS every lesson has I think its called "Function question" underneath is a question. Thoes questions are for the dba. There are questions at the top of every page of every lesson. Look uo the answers for them! or use the notes most of the time they give you answer to them already. I hope that helps of you have any other questions feel free to ask <3

6 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

2 years ago

How can you use the point-slope form of a line when only given two points?

pshichka [43]

Answer:

See below

Step-by-step explanation:

Point -slope form is:

y - y1 = m(x - x1)

By having one set of coordinates and knowing the value of the slope (m in the equation) you can use the form.

How to find the slope?

Here you need 2 points: Point 1 with coordinates x1, y1 and point 2 with the coordinates x2, y2

Use the formula:

m = (y2-y1)/(x2-x1)

Example:

m = -5, and one of the points has coordinates (9, 2) then point-slope form is:

y - 2 = -5(x - 9)

3 0

3 years ago

Find the vertical and horizontal asymptotes, domain, range, and roots of f (x) = -1 / x-3 +2.

gladu [14]

Answer:

Vertical asymptote: $x=3$

Horizontal asymptote: $f(x) =2$

Domain of f(x) is all real numbers except 3.

Range of f(x) is all real numbers except 2.

Step-by-step explanation:

Given:

Function:

$f (x) = -\dfrac{1 }{ x-3} +2$

One root, $x = 3.5$

To find:

Vertical and horizontal asymptote, domain, range and roots of f(x).

Solution:

First of all, let us find the roots of f(x).

Roots of f(x) means the value of x where f(x) = 0

$0= -\dfrac{1 }{ x-3} +2\\\Rightarrow 2= \dfrac{1 }{ x-3}\\\Rightarrow 2x-2 \times 3=1\\\Rightarrow 2x=7\\\Rightarrow x = 3.5$

One root, $x = 3.5$

Domain of f(x) i.e. the values that we give as input to the function and there is a value of f(x) defined for it.

For x = 3, the value of f(x) $\rightarrow \infty$

For all, other values of $x$ , $f(x)$ is defined.

Hence, Domain of f(x) is all real numbers except 3.

Range of f(x) i.e. the values that are possible output of the function.

f(x) = 2 is not possible in this case because something is subtracted from 2. That something is $\frac{1}{x-3}$ .

Hence, Range of f(x) is all real numbers except 2.

Vertical Asymptote is the value of x, where value of f(x) $\rightarrow \infty$ .

$-\dfrac{1 }{ x-3} +2 \rightarrow \infty$

It is possible only when

$x-3=0\\\Rightarrow x=3$

$\therefore$ vertical asymptote: $x=3$

Horizontal Asymptote is the value of f(x) , where value of x $\rightarrow \infty$ .

$x\rightarrow \infty \Rightarrow \dfrac{1 }{ x-3} \rightarrow 0\\\therefore f(x) =-0+2 \\\Rightarrow f(x) =2$

$\therefore$ Horizontal asymptote: $f(x) =2$

Please refer to the graph of given function as shown in the attached image.

5 0

3 years ago