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2 years ago

What is 2 3/8 in a improper fraction?

Mathematics

2 answers:

PIT_PIT [208]2 years ago

8 0

The answer to that is 19/8

FromTheMoon [43]2 years ago

4 0

Answer:

See below a shorter way to convert 2 3 8 to an improper fraction. Step 1: Multiply the whole number part (2) by the denominator (8). Step 2: Add the product from Step 1 (16) to the numerator (3). Step 3: Write that result (19) above the denominator.

Step-by-step explanation:

hope this helps you if it does plz mark brainiest

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Which expression is equivalent to r^9/r^3

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$\cfrac{r^9}{r^3} =r^{9-3}=r^6$

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Find the values of x and y. 48° (5x – 17) X = and y =

likoan [24]

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2 years ago

I need help with this question

Novay_Z [31]

Answer:

$\frac{\sqrt{3} - 1}{2\sqrt{2}}$

$\frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

$- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Step-by-step explanation:

Given $\frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6}$

(A) $sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4} + \frac{\pi}{6})$

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2}$

$$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

(B) Cos(A + B) = CosAcosB - SinASinB

$cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}})$

$\implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2}$

$\implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6})$

$$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

From the above obtained values this can be calculated and the value is $- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ .

3 0

2 years ago

Need some help please and thanks

Elina [12.6K]

Answer:

answer is c

Step-by-step explanation:

x-15<4 x-15>-4

+15 +15

------------------------------------

x<19 x>11

11<x<19

3 0

3 years ago

Please Help. Will give brainliest.

Keith_Richards [23]

Answer:

mDBC = 341 °

Step-by-step explanation:

Arc DBC is almost the whole circle, if you were to exclude the degree measure of minor arc DC; which is, in other words m∠ DPC

Now by Vertical Angles Theorem:

m ∠ DPC = m∠ APB, ⇒ provided BD, and AC are diameters

m∠ DPC = 19°

This would mean ⇒

mDBC = 360 - m∠ DPC,

mDBC = 360 - 19,

mDBC = 341 degrees ( ° )

3 0

2 years ago