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2 years ago

What is the product of 5⁹ and 5 to the twelfth power

Mathematics

1 answer:

Lemur [1.5K]2 years ago

7 0

Answer:

19,073,486,328,125

Step-by-step explanation:

$5x^9 * 5*5*5*5*5*5*5*5*5*5 = 19,073,486,328,125$

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If m∠A = 56°, m∠B = 60°, and c = 8, what is the measure of side length b?

Alekssandra [29.7K]

Using the law of sines:

$\frac{b}{sin(B)}=\frac{c}{sin(C)}$

Solve for b:

$\begin{gathered} b=\frac{c\cdot sin(B)}{sin(C)} \\ so: \\ b=\frac{8\cdot sin(60)}{sin(64)} \\ b\approx7.708 \end{gathered}$

Answer:

7.708

5 0

1 year ago

What is 16 to the one billionth power

qwelly [4]

The answer Its 4.5281 ....etc

3 0

3 years ago

Find the hypotenuse, , of the triangle

Irina18 [472]

Answer:

hypoténuse = 15 in

b ² = 15² - 9²

b ² = 144

b = √144 = 12

Perimeter = 15 + 9 + 12 = 36 inches

Step-by-step explanation:

5 0

2 years ago

A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a

natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, . . . , x₂₃ be the actual data observed by the student

The sample means = x₁ + x₂ + . . . , x₂₃ / 23

$= \frac{x_1 +x_2 +...x_2_3}{23}$

= 2.3hr

⇒ $\sum xi =2.3 \times 23 = 52.9hrs$

let x₁ , x₂, . . . , x₂₃ arranged in ascending order

Then x₂₃ was 10 and has been changed to 14

i.e x₂₃ increase to 4

Sample mean $= \frac{x_1 +x_2 +...x_2_3}{23}$

$\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47$

therefore, the new sample mean is 2.47

2) For the old data set

the median is $x_1_2(th)$ values

$[\frac{n +1}{2} ]^t^h value$

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = $[{x_1_2]^t^h value$ remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

$=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)$

The new sample standard sample deviation is calculated as

$= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)$

Now, when we compare (1) and (2) the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)² ∑ (xi - 2.47)²

Therefore , the standard deviation increases

6 0

3 years ago

It is not possible to prove one pair of triangles congruent and then use their congruent corresponding parts to prove another pa

borishaifa [10]

Answer:

true

The wording does not quite mean anything,

but what I think was meant to ask is

"if we use some parts of two triangles to prove they are congruent,

can we then use that to prove that

a pair of corresponding parts not used before are congruent?"

The answer is

Yes, of course,

Corresponding Parts of Congruent Triangles are Congruent,

which teachers usually abbreviate as CPCTC.

For example, if we find that

side AB is congruent with side DE,

side BC is congruent with side EF, and

angle ABC is congruent with angle DEF,

we can prove that triangles ABC and DEF are congruent

by Side-Angle-Side (SAS) congruence.

We then, by CPCTC, can conclude that other pairs of corresponding parts are congruent:

side AB is congruent with side DE,

angle BCA is congruent with angle EFD, and

angle CAB is congruent with angle FDE.

It was possible (by CPCTC) to prove those last 3 congruence statements,

after proving the triangles congruent.

The expected answer is FALSE.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers