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3 years ago

6

Help please I cant get this right

1 answer:

ruslelena [56]3 years ago

8 0

if it were me I would start by inputting 3 into the equation and then trying to simplify it

$\sqrt{3^2+6*3+9}$

$\sqrt{9+18+9}$

$\sqrt{36}$

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Answer truthfully pleaseeeeee

alexandr1967 [171]

Answer:

1/4

Step-by-step explanation:

i took the quiz

3 0

2 years ago

Find the 27th term in the arithmetic sequence of 59,56,53,50

statuscvo [17]

solution:

$A_{n}=59-3(n-1)\\where, n=1,2,3,.....\\A1=59\\A2=56\\A3=53\\A_{n}=a+(n-1)dA27=59+(27-1)3\\=59-78\\=-19\\A27 term is -19$

5 0

3 years ago

A random sample of 7 fields of durum wheat has a mean yield of 29.8 bushels per acre and standard deviation of 3.62 bushels per

meriva

Answer:

CI = 29.8 ± 3.53

Critical value is z = 2.58

Step-by-step explanation:

First of all let's find margin of error. It is given by the formula;

ME = zσ/√n

We are given;

Standard deviation; σ = 3.62

Sample size; n = 7

Mean; x¯ = 29.8

Now, z-value for 99% Confidence level is 2.58

Thus;

ME = (2.58 × 3.62)/√7

ME = 3.53

CI is written as;

CI = x¯ ± ME

CI = 29.8 ± 3.53

Critical value is z = 2.58

5 0

2 years ago

The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?

11111nata11111 [884]

we have

$A(-2, 2),B(6, 2),C(0, 8)$

see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

$P=AB+BC+AC$

and

the area of the triangle is equal to

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB\\heigth=DC$

we know that

The distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step $1$

<u>Find the distance AB</u>

$A(-2, 2),B(6, 2)$

Substitute the values in the formula

$d=\sqrt{(2-2)^{2}+(6+2)^{2}}$

$d=\sqrt{(0)^{2}+(8)^{2}}$

$dAB=8\ units$

Step $2$

<u>Find the distance BC</u>

$B(6, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-6)^{2}}$

$d=\sqrt{(6)^{2}+(-6)^{2}}$

$dBC=6\sqrt{2}\ units$

Step $3$

<u>Find the distance AC</u>

$A(-2, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0+2)^{2}}$

$d=\sqrt{(6)^{2}+(2)^{2}}$

$dAC=2\sqrt{10}\ units$

Step $4$

<u>Find the distance DC</u>

$D(0, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-0)^{2}}$

$d=\sqrt{(6)^{2}+(0)^{2}}$

$dDC=6\ units$

Step $5$

<u>Find the perimeter of the triangle</u>

$P=AB+BC+AC$

substitute the values

$P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units$

$P=22.81\ units$

therefore

The perimeter of the triangle is equal to $22.81\ units$

Step $6$

<u>Find the area of the triangle</u>

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB=8\ units\\heigth=DC=6\ units$

substitute the values

$A=\frac{1}{2}*8*6$

$A=24\ units^{2}$

therefore

the area of the triangle is $24\ units^{2}$

4 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BE%7D%7B%20%5Csqrt%7BR%20%7B%7D%5E%7B2%7D%20%2B%20W%20%7B%7D%5E%7B2%7D%20L%

AleksandrR [38]

Answer:

R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Step-by-step explanation:

Squaring both sides of equation:

I^2 = (ER)^2/(R^2 + (WL)^2)

<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)

<=>(ER)^2 - (IR)^2 = (IWL)^2

<=> R^2(E^2 - I^2) = (IWL)^2

<=> R^2 = (IWL)^2/(E^2 - I^2)

<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Hope this helps!

7 0

3 years ago