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3 years ago

I would be double blessed if you could answer this please n thank you

Mathematics

1 answer:

valina [46]3 years ago

3 0

X = 8
m ∠GDH = 126°
m ∠FDH = 160°
m ∠FDE = 137°

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Which inverse operation should be performed first?: m = 7 = 32.

Mumz [18]

The answer is multiplication

7 0

3 years ago

I need help plz! If anyone knows, I would truly appreciate it! Thank uuu!!!!^^

swat32

Answer:

pic1 a pic2 d

Step-by-step explanation:

i did this

8 0

3 years ago

Read 2 more answers

(cosxtanx-tanx+2cosx-2)/tanx+2=cosx-1<br> can someone show me how to prove this?

uysha [10]

Answer:

See detail below.

Step-by-step explanation:

A word of caution before getting to the actual problem: I believe there is an important set of brackets missing in the original post. The expression on the left hand side should be:

(cosxtanx-tanx+2cosx-2)/(tanx+2)

Without the brackets, it is left unclear whether the denominator is just tanx or tanx+2. I recommend to use brackets wherever any doubt could arise.

Now to the actual problem: \we can make the following transformations on the left hand side:

$\frac{\cos x \tan x + 2\cos x -2}{\tan x +2}=\frac{\cos x \frac{\sin x}{\cos x} + 2\cos x -2}{\frac{\sin x +2\sin x}{\cos x} }=\\=\frac{\cos x \sin x - \sin x + 2\cos^2 x - 2\cos x}{\sin x + 2\cos x}=\\=\frac{\cos x \sin x +2\cos^2 x }{\sin x + 2\cos x}-1=\\=\cos x -1$

which is shown to be the same as the right hand side, which was to be shown.

7 0

3 years ago

Given two independent random samples with the following results: n1=8x‾1=186s1=33 n2=7x‾2=171s2=23 Use this data to find the 90%

KonstantinChe [14]

Answer:

The point of estimate for the true difference would be:

$186-171= 15$

And the confidence interval is given by:

$(186-171) -1.77 \sqrt{\frac{33^2}{8} +\frac{23^2}{7}}= -10.753$

$(186-171) +1.77 \sqrt{\frac{33^2}{8} +\frac{23^2}{7}}= 40.753$

Step-by-step explanation:

For this case we have the following info given:

$\bar X_1 = 186$ the sample mean for the first sample

$\bar X_2 = 171$ the sample mean for the second sample

$s_1 =33$ the sample deviation for the first sample

$s_2 =23$ the sample deviation for the second sample

$n_1 = 8$ the sample size for the first group

$n_2 = 7$ the sample size for the second group

The confidence interval for the true difference is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

We can find the degrees of freedom are given:

$df = n_1 +n_2 -2 =8+7-2= 13$

The confidence level is given by 90% so then the significance would be $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ we can find the critical value with the degrees of freedom given and we got: