Ask question

Ask question

All categories

3 years ago

12

Find the value for x? X-5+5x-12=25

1 answer:

levacccp [35]3 years ago

5 0

X-5+5x-12=25
x+5x=25+5+12
6x=42
x=42/6
x=7

You might be interested in

Please answer ASAP will mark brainilest

Otrada [13]

Answer:

1

Step-by-step explanation:

Change of Y is 0 and change of x is any number.

0/any number = 0

6 0

3 years ago

Write an equation of the line passing through the given point (6,-7) and having the given slope m=-9. Write the final answer in

Ede4ka [16]

Y-(-7)=-9(x-6)
Y+7= -9x+54
Subtract 7 from both sides
Y= -9x+47

3 0

3 years ago

Can someone help me find the diameter? <br><br> also what is a diameter even?

lukranit [14]

Answer:

24

Step-by-step explanation:

area of circle = πr²

πr² = 144π

Cancel out π

r² = 144

r = √144

r = 12

diameter = 2 x radius

diameter = 2 x 12

diameter = 24m

* diameter is the "width of a circle", the distance from the widest points of the circle

3 0

2 years ago

Two competing headache remedies claim to give fast-acting relief. An experiment was performed to compare the mean lengths of tim

DaniilM [7]

Answer:

$t = 0.875$

Step-by-step explanation:

Given

<u>Brand A</u> <u>Brand B</u>

$n_ 1= 12$ $n_2 = 12$

$\bar x_1 = 21.8$ $\bar x_2 = 18.9$

$\sigma_1 = 8.7$ $\sigma_2 = 7.5$

Required

Determine the test statistic (t)

This is calculated as:

$t = \frac{\bar x_1 - \bar x_2}{s\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

Calculate s using:

$s = \sqrt{\frac{(n_1-1)*\sigma_1^2+(n_2-1)*\sigma_2^2}{n_1+n_2-2}}$

The equation becomes:

$s = \sqrt{\frac{(12-1)*8.7^2+(12-1)*7.5^2}{12+12-2}}$

$s = \sqrt{\frac{1451.34}{22}}$

$s = \sqrt{65.97}$

$s = 8.12$

So:

$t = \frac{\bar x_1 - \bar x_2}{s\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

$t = \frac{21.8 - 18.9}{8.12 * \sqrt{\frac{1}{12} + \frac{1}{12}}}$

$t = \frac{21.8 - 18.9}{8.12 * \sqrt{\frac{1}{6}}}$

$t = \frac{21.8 - 18.9}{8.12 * 0.408}}$

$t = \frac{2.9}{3.31296}}$

$t = 0.875$

3 0

3 years ago

Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len

Rama09 [41]

The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

$\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt$

In this case, we have

<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> ) ==> d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )

<em>y(t)</em> = 5 - 2<em>t</em> ==> d<em>y</em>/d<em>t</em> = -2

and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then

$\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt$

$=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt$

$=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt$

$=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt$

$=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}$

5 0

2 years ago