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2 years ago

Simplify the expression 5a – 2b – 3 + 2b – 6a A. –4 B. 2a – 4b C. –a – 3 D. a – 3

Mathematics

1 answer:

ad-work [718]2 years ago

8 0

Answer:

C. -a-3

Step-by-step explanation:

Group like terms:

5a and -6a

-2b and +2b

-3

5a - 6a equals -a (-1a, but you don't write the 1)

-2b and + 2b = 0

Now you have -a and -3.

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Find the slope given the two points. (2, -6) and (2, 3)

kifflom [539]

Answer: Undefined

The x coordinates are the same, so a vertical line forms. All vertical lines have undefined slopes.

We can see it through the slope formula

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

m = (3-(-6))/(2-2)

m = (3+6)/(2-2)

m = 9/0

We cannot divide by zero, so the result is undefined.

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3 years ago

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Please help on this question for me!

Gnoma [55]

Answer:

parallelograms. angles

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3 years ago

A random sample of 64 observations produced a mean value of 84 and standard deviation of 5.5. The 90% confidence interval for th

ASHA 777 [7]

Answer: The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

Step-by-step explanation:

When population standard deviation is not given ,The confidence interval population proportion is given by ( $\mu$ ):-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where n= Sample size.

s= Sample standard deviation

$\overline{x}$ = sample mean

t* = Critical t-value (Two-tailed)

As per given , we have

$\overline{x}=84$

n= 64

Degree of freedom : df = n-1=63

s= 5.5

Significance level : $\alpha=1-0.90=0.1$

Two-tailed T-value for df = 63 and $\alpha=1-0.90=0.1$ would be

$t_{\alpha/2,df}=t_{0.05,63}=1.669$ (By t-distribution table)

i.e. t*= 1.669

The 90% confidence interval for the population mean μ would be

$84\pm (1.669)\dfrac{5.5}{\sqrt{64}}$

$=84\pm (1.669)\dfrac{5.5}{8}$

$\approx84\pm 1.15$

$=(84-1.15,\ 84+1.15)=(82.85,\ 85.15)$

∴ The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

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4 years ago

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DanielleElmas [232]

Answer:

320km = 12L

296 km = xL

x = (296 × 12)/320

x = 11.1 liters

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3 years ago

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Integrate e^x(sin(x) cos(x))

Karo-lina-s [1.5K]

$I=\int e^x(\sin(x)\cos(x))dx=\int e^x(\frac{1}{2}\sin(2x))dx=\frac{1}{2}\int e^x\sin(2x)dx$

$\text{If }u=\sin(2x)\to du=2\cos(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
\text{Using }\int u\,dv=uv-\int v\,du:\\\\
I=\frac{1}{2}(e^x\sin(2x)-\int e^x(2\cos(2x))dx)\\\\
2I=e^x\sin(2x)-2\underbrace{\int e^x\cos(2x)dx}_{I_2}$

Looking for $I_2$ :

$\text{If}~u=\cos(2x)\to du=-2\sin(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
I_2=e^x\cos(2x)-\int e^x(-2\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(2\sin(x)\cos(x))dx\\\\ I_2=e^x\cos(2x)+4\int e^x(\sin(x)\cos(x))dx=e^x\cos(2x)+4I$

Replacing:

$2I=e^x\sin(2x)-2I_2\iff\\\\2I=e^x\sin(2x)-2(e^x\cos(2x)+4I)\iff\\\\
2I=e^x\sin(2x)-2e^x\cos(2x)-8I\iff\\\\
10I=e^x\sin(2x)-2e^x\cos(2x)\iff\\\\
I=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))\\\\
\boxed{\int e^x(\sin(x)\cos(x))dx=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))+C}$

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4 years ago