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3 years ago

Use a calculator to solve this problem: 28 + 27 × 36

2 answers:

8 0

For this case we have that according to the order of algebraic operations PEMDAS, the multiplications have priority over the addition and subtraction, then, we have the following expression:

$28 + 27 * 36 =\\28 + 972 =\\1000$

Thus, the value of the expression is 1000.

Answer:

$28 + 27 * 36 = 1000$

julsineya [31]3 years ago

6 0

Answer:

Final answer is 28 + 27 × 36 = 1000.

Step-by-step explanation:

Given expression is 28 + 27 × 36.

Now we need to simplify that using calculator.

There are two operations +(Addition) and x (Multiplication).

According to the order of operation, we should multiply first then add.

So 27x36 = 972.

Now add it with 28.

28 + 972 = 1000

Hence final answer is 28 + 27 × 36 = 1000.

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Find the measure of XCY. PLEASE HELP ME AND SHOW WORK

Dovator [93]

Answer:

Measure of angle XCY is 60°.

Step-by-step explanation:

"Measure of all inscribed angles by the same arc in one segment of the circle are equal"

Using this property in the given circle,

m∠XWY = m∠XCY

60x = 61x - 1

61x - 60x = 1

x = 1

By substituting the value of 'x' in the measure of ∠XCY,

m∠XCY = 61x - 1

= 61(1) - 1

= 61 - 1

= 60°

Therefore, measure of angle XCY is 60°.

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2 years ago

A rocket generates a net force (F) of 2,050,000 newtons, and the rocket mass (m) is 40,000 kilograms

SVEN [57.7K]

From the given equation above, F = ma, acceleration may be calculated by slightly modifying the equation into a = F / m. Substituting the known values for force and mass,

a = 2,050,000 N / 40,000 kg = 51.25 m/s²

Thus, the acceleration achieved is 51.25 m/s².

6 0

2 years ago

How many groups of Two-thirds are in Three-fifths?

Vanyuwa [196]

Answer:

its D

Step-by-step explanation:

Did the test

8 0

3 years ago

Read 2 more answers

Given the recursive formula find the first 4 terms for each. (not multiple choice) (MAFS.912.F-BF.1.3) A. An=an-1 3 when a1=5 B.

Neko [114]

Answer:

<h3>See explanations below</h3>

Step-by-step explanation:

1) Given the recursive function An=an-1 + 3 when a1 = 5, we are to find the first four terms;

First term a1 = 5

a2 = a1 +3

a2 = 5 + 3

a2 = 8

a3 = a2 + 3

a3 = 8+3

a3 = 11

a4 = a3 + 3

a4 = 11 + 3

a4 = 14

The first four terms are 5, 8, 11 and 14

2) For the recursive function An=an-1 + 2/3 when a1 = 1

a2 = a1 + 2/3

a2 = 1 + 2/3

a2 = 5/3

a3 = a2 + 2/3

a3 = 5/3 + 2/3

a3 = 7/3

a4 = a3 + 2/3

a4 = 7/3 + 2/3

a4 = 9/3

a4 = 3

Hence the first four terms of the sequence are 2/3, 5/3, 7/3, 3

3) For the recursive function An=an-1 + 12 when a1=30

a2 = a1 + 12

a2 = 30 + 12

a2 = 42

a3 = a2 +12

a3 = 42 + 12

a3 = 54

a4 = a3 + 12

a4 = 54+12

a4 = 66

Hence the first four terms of the sequence are 30, 42, 54, 66

3 0

2 years ago

I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

8 0

1 year ago