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2 years ago

S = 2B + L. What do S, B, and L stand for?

Mathematics

1 answer:

notsponge [240]2 years ago

7 0

Hi Jordyn! I think S is segment, B is base, and L is length. I hope this helped!

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Pls help Im very stuck on this and don’t understand

Lady bird [3.3K]

Answer:

C. 85 degrees

Step-by-step explanation:

Step 1: Set up equation

x + 57 + 96 + x = 131

Step 2: Solve for <em>x</em>

2x + 153 = 131

2x = -22

x = -11

Step 3: Find m∠RQU

m∠RQU = 96 + <em>x</em>

m∠RQU = 96 - 11

m∠RQU = 85°

8 0

3 years ago

PLEASE HELP

AnnZ [28]

Answer: 7

Step-by-step explanation:

1. multiply both sides by -6 so the -6 on the right side could cancel out and the left side would be -6: 1= a-13/-6 ==> -6=a-13

2. add both sides by 13 so -13 on the right side could cancel out and the left side would be -6+13: -6=a-13 ==> -6+13=a-13+13

3. you know -6+13 equals 7 right? so the answer is a=7

5 0

2 years ago

Read 2 more answers

An article reported that for a sample of 55 kitchens with gas cooking appliances monitored during a one-week period, the sample

Sati [7]

Answer:

Step-by-step explanation:

Given that

sample size n = 55: x bar = 654.16 and s = sample sd = 162.34

Std error = 162.34/sqrt 55 = 21.889

For 95% CI we can use t critical value as population std dev is not known.

df = 54

t critical = 2.004

Margin of error = 2.004 *21.889 = 43.866

Confidence interval lower bound = 654.16-43.866 =610.294

Upper bound = 654.16+43.866=698.026

Confidence interval rounded off at 95% = (610.29, 698.23)

8 0

2 years ago

What is 10x5+24+9?? i need help

ki77a [65]

It is 84. multply 10 and 5, then add the rest. hope i could help!

7 0

3 years ago

Read 2 more answers

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

2 years ago