A triangle has angle measures of 44°,(3x+8), and (4c-12)°. Classify the triangle by its angles

Condense the expression to the logarithm of a single quantity.<br><br>

vladimir1956 [14]

Answer:

$\log _3\left(\frac{x^{\frac{1}{2}}}{\left(y+8\right)^2}\right)$

Step-by-step explanation:

-> Apply log rules

$\log _3\left(x^{\frac{1}{2}}\right)-2\log _3\left(y+8\right)$

->

$\log _3\left(x^{\frac{1}{2}}\right)-\log _3\left(\left(y+8\right)^2\right)$

-> Apply basic math rules

$\log _3\left(\frac{x^{\frac{1}{2}}}{\left(y+8\right)^2}\right)$

7 0

1 year ago

Dan says he ran twice as far as Karim. Give three possibilities for the distances each could have run.

slamgirl [31]

These people never answer thesee

7 0

2 years ago

I need help with 5 1\2× 3 1\4 =?

sattari [20]

5 1/2 * 3 1/4

1 + 5 . 2/2 * 1 + 3 . 4/4 = 11/2 * 13/4 = 11 . 13/2 . 4 = 143/8 = 7 + 17 . 8/8 = 17 7/8

= 17 . 875

Detailed explanation of solution:

5 1/2 * 3 1/4

Transform the first mix number to improper fraction:

5 1/2 = 1 + 5 . 2/2 = 11/2

Transform the second mix number to improper fraction:

3 1/4 = 1 + 3. 4/4 = 13/4

Multiplying two fractions:

11/2 * 13/4 = 11 . 13/2 . 4 = 143/8

Since the numerator is greater than the denominator, We convert the improper fraction to mix fraction:

143/8 = 7 + 17.8/8

= 17 7/8 (Decimal: 17. 875)

Hope that helps!!!!!! ( Answer: 17 7/8, (Decimal: 17.875)

3 0

3 years ago

What is the correct answer to this?

ra1l [238]

Answer:

33.4933 ≈ 33.5 mm³

Step-by-step explanation:

Given that,

→ Radius (r) = 2 mm

Formula we use,

→ 4/3 πr³

Let's solve for the volume of sphere,

→ 4/3 πr³

→ 4/3 × 3.14 × 2 × 2 × 2

→ 4/3 × 3.14 × 8

→ 4/3 × 25.12

→ 100.48/3

→ 33.4933

→ [ ≈ 33.5 mm³ ]

Hence, volume of sphere is 33.5 mm³.

7 0

1 year ago

Read 2 more answers

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the

Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 3.6

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3.6 - 3.6}{0.4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 4.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 3

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3 - 3.6}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

$Z = \frac{X - \mu}{\sigma}$

$1.75 = \frac{X - 3.6}{0.4}$

$X - 3.6 = 0.4*1.75$

$X = 4.3$

They last at least 4.3 minutes

7 0

2 years ago