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3 years ago

Multiply and simplify 2/7 and 3/5

Mathematics

2 answers:

lana66690 [7]3 years ago

7 0

Answer:

Step-by-step explanation:

2/7*3/5=

2x3=6

7x5=35

6/35

Simplify...

It can't be simplified

gladu [14]3 years ago

4 0

Answer: 6/35

Step-by-step explanation:

2/7 * 3/5 = (2 * 3) / (7 * 5) = 6/35

And that is in simplest form.

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-2x + 2y = -4

kaheart [24]

Answer: x=2

Step-by-step explanation: refer to image

6 0

3 years ago

HELP PLZ WILL MARK U AS BRAINLIEST

Alla [95]

Answer:

Difference in volume between the two vinyl records = 3.872 in³

Step-by-step explanation:

Complete Question

2 Columbia records unveiled the LP (a vinyl record) in the Waldorf Astoria on June 18, 1948, in two formats: 10 inches in diameter, matching that of 78 rpm singles, and 12 inches in diameter. If the thickness of one vinyl record is 0.112 in, then determine the difference in volumes between the 10 inch and 12 inch records.

Each of the vinyl records will be modeled as a cylinder

Volume of a cylinder = πr²h

For the 12 inches diameter records:

r = (12/2) = 6 inches

h = thickness = 0.112 inches

Volume = π × (6²) × 0.112 = 12.672 in³

For the 10 inches diameter records:

r = (10/2) = 5 inches

h = thickness = 0.112 inches

Volume = π × (5²) × 0.112 = 8.8 in³

Difference in volume between the two vinyl records = 12.672 - 8.8 = 3.872 in³.

Therefore, the difference in volume between the two vinyl records = 3.872 in³.

Have a nice day!

3 0

3 years ago

Plz with steps .. it's very hard can anyone plz

liubo4ka [24]

Answer:

Step-by-step explanation:

$\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a} \qquad f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a} \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} }{\dfrac{1}{2\sqrt{x}} }\\\\$

$\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}} =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\$

7 0

3 years ago

Which number satisfies the inequality? 3/11 < x <sqrt 0.25<br><br>A. 30%<br>B. 7/9

olya-2409 [2.1K]

Wait what do you mean

4 0

3 years ago

Explain why a geometric series with a ratio between zero and one converges and how you find the sum.

Alex Ar [27]

Let $S_n$ be the $n$ th partial sum of a geometric sequence with common ratio $r$ and first term $a$ . So the sequence is $\{a,ar,ar^2,\cdots\}$ , and $S_n$ is

$S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}$

Multiplying both sides by $r$ gives

$rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^n$

and subtracting $rS_n$ from $S_n$ gives

$S_n-rS_n=a+(ar-ar)+(ar^2-ar^2)+\cdots+(ar^{n-1}-ar^{n-1})-ar^n$
$(1-r)S_n=a(1-r^n)$
$S_n=a\dfrac{1-r^n}{1-r}$

If $0$ , then $r^n\to0$ as $n\to\infty$ and so the sum approaches

$\displaystyle\lim_{n\to\infty}S_n=\dfrac a{1-r}$

6 0

3 years ago

Read 2 more answers