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3 years ago

Given: Tangents LA and LB m∠AOB=110° Find: m∠ALO.

Mathematics

1 answer:

Ad libitum [116K]3 years ago

7 0

Answer:

35°

Step-by-step explanation:

From any point outside a circle, the two tangents to that circle are always congruent to each other. This can be proven by taking the two triangles ΔLAO and ΔLBO and prove that they are congruent. From the upper triangle AO is congruent to BO because AO = BO = Radius. Also, LO is the same length for both triangles, therefore two equal sides. Moreover, the measure of an angle between the tangent to a point of the circle and the radius going from the center to that point equals 90°, so ∠LAO = ∠LBO = 90°. So these two triangles ΔLAO and ΔLBO are congruent by SSA postulate.

If m∠AOB=110°, then m∠AOL = 110°/2 =55°. Finally, the angles of every triangle always measures 180°, therefore:

m∠ALO = 180°-55°-90 = 35°

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Answer:

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Step-by-step explanation:

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Convert point slope form to slope intercept form by simplifying and rearranging.

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3 years ago

The length of an object is measured as 34.8 centimeters. What is this measurement in scientific notation?

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Step-by-step explanation:

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2 years ago

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aleksandr82 [10.1K]

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2 years ago

Solve for

faust18 [17]

Given equation : n(17+x)=34x−r.

We need to solve it for x.

Distributing n over (17+x) on left side, we get

17n + nx = 34x - r.

Adding r on both sides, we get

17n+r + nx = 34x - r+r.

17n + r + nx = 34x.

Subtracting nx from both sides, we get

17n + r + nx-nx = 34x-nx

17n + r = 34x -nx.

Factoring out gcf x on right side, we get

17x + r = x(34-n).

Dividing both sides by (34-n), we get

$\frac{(17x + r)}{(34-n)} = \frac{x(34-n)}{(34-n)}$

$\frac{(17x + r)}{(34-n)} = x$

<h3>Therefore, final answer is $x=\frac{(17x + r)}{(34-n)}.$ </h3>

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3 years ago

Read 2 more answers

The bad debt ratio for a financial institution is defined to be the dollar values of loans defaulted divided by the total dollar

Nimfa-mama [501]

Answer:

(a) NULL HYPOTHESIS, $H_0$ : $\mu \leq$ 3.5%

ALTERNATE HYPOTHESIS, $H_1$ : $\mu$ > 3.5%

(b) We conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Step-by-step explanation:

We are given that a random sample of seven Ohio banks is selected.The bad debt ratios for these banks are 7, 4, 6, 7, 5, 4, and 9%.The mean bad debt ratio for all federally insured banks is 3.5%.

We have to test the claim of Federal banking officials that the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

(a) Let, NULL HYPOTHESIS, $H_0$ : $\mu \leq$ 3.5% {means that the the mean bad debt ratio for Ohio banks is less than or equal to the mean for all federally insured banks}

ALTERNATE HYPOTHESIS, $H_1$ : $\mu$ > 3.5% {means that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks}

The test statistics that will be used here is One-sample t-test;

T.S. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean debt ratio of Ohio banks = 6%

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 1.83%

n = sample of banks = 7

So, test statistics = $\frac{6-3.5}{\frac{1.83}{\sqrt{7} } }$ ~ $t_6$

= 3.614

(b) Now, at 1% significance level t table gives critical value of 3.143. Since our test statistics is more than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Hence, Federal banking officials claim was correct.

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3 years ago