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3 years ago

Use multiplication to find three equivalent ratios 11 : 9

Mathematics

1 answer:

dlinn [17]3 years ago

5 0

11 : 9

×2 ×2

22 : 18

11 : 9

×3 ×3

33 : 27

11 : 9

×4 ×4

44 : 36

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.092 -.15-.19= how do I solve

photoshop1234 [79]

Line up the decimal point to each number in order from left to right and subtract to get the difference which should be -0.248

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3 years ago

There were 1372 million fewer barrels of crude oil produced in country A in 2009 than there were in country B. Find the number o

snow_lady [41]

Easy, the answer is 1,588 million barrels of oil produced in country B... just add 1372 to 216...

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4 years ago

The tangent of an angle is 3.4. What is the measure of the angle to the nearest tenth? 52.9° 73.6° 99.8° 68.2°

ryzh [129]

Arctan(3.4) ≈ 73.6°. You can view the 3.4 as the slope of a line. Then it is clear the the 52.9 and 99.8 can't be right. If you have to do this without a calculator, it can be quite tricky...

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4 years ago

What is the range for the following function? y=1/(x+2)+3

LUCKY_DIMON [66]

Answer:

$\large \boxed{\text{C) }\{y: y \in \mathbb{R}, y \ne 3\} }$

Step-by-step explanation:

The range is the spread of the y-values (minimum to maximum distance travelled).

The graph of your function is a hyperbola shifted two units left and three units up from the origin.

There is a vertical asymptote at x = -2, so y does not exist when x = -2. However,

$\displaystyle \lim_{x \rightarrow -{2}^{+}}f(x) = \lim_{x \rightarrow -{2}^{+}}\left (\dfrac{1}{x+2}+3 \right ) = 0 + 3 = 3\\\\\lim_{x \rightarrow -{2}^{-}}f(x) = \lim_{x \rightarrow -{2}^{-}}\left (\dfrac{1}{x+2}+3 \right ) = 0 + 3 = 3$

Since the limit from either side is the same,

$\displaystyle \lim_{x \rightarrow -{2}}f(x) = 3$

The graph below shows the asymptotes of your function.

Thus. y can take any value except 3.

In set builder notation, the range is

$\large \boxed{\mathbf{\{y: y \in \mathbb{R}, y \ne 3\} }}$

4 0

3 years ago

Find the rectangular coordinates of the point. (-5, -180°)

o-na [289]

$\bf \begin{array}{rlclll}
(-5&,&-180^o)\\
\uparrow &&\uparrow \\
r&&\theta 
\end{array}\quad 
\begin{cases}
x=rcos(\theta )\\
\qquad -5\cdot cos(-180^o)\\
\qquad -5\cdot -1\\
\qquad 5\\
----------\\
y=rsin(\theta )\\
\qquad -5\cdot sin(-180^o)\\
\qquad -5\cdot 0\\
\qquad 0
\end{cases}\implies (5,0)$

4 0

3 years ago