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dem82 [27]
3 years ago
14

Dilate the segment about the origin by a scale factor of 1 /2 What are the coordinates of B'? please help

Mathematics
2 answers:
labwork [276]3 years ago
7 0

Answer: simply the answer is b

Step-by-step explanation:

Marina86 [1]3 years ago
5 0

Answer:

(1,-2) edge2020

Step-by-step explanation:

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Jess has 60 ounces of an alloy that is 40% gold. How many ounces of pure gold must be added to this alloy to create a new alloy
tia_tia [17]

Answer:

84 ounces of pure gold

Step-by-step explanation:

Jess has 60 ounces of an alloy that is 40% gold. How many ounces of pure gold must be added to this alloy to create a new alloy that is 75% gold?

Pure gold = 100% gold

Let the number of ounces of pure gold = x

Hence, we have the equation

40% × 60 ounces + 100%× x ounces = 75%(60 + x)ounces

= 0.4 × 60 + 1x = 0.75(60 + x)

= 24 + x = 45 + 0.75x

Collect like terms

x - 0.75x = 45 - 24

0.25x = 21

x = 21/0.25

x = 84 ounces

Therefore, we need 84 ounces of pure gold

6 0
3 years ago
Name 2 pair of interior angles
ivanzaharov [21]
Adjacent interior angels
8 0
3 years ago
Zack deposited $1,200 in a savings account that paid 7.75% simple
Eddi Din [679]
<span>The correct answer is: Option (C) $1393.21

Explanation:
At the beginning of the Second year, the total balance would be:

$1200 + ($1200 * 7.75 / 100) = $1293

At the beginning of the Third year, the total balance would be:
$1293 + ($1293 * 7.75 / 100) = $1393.21 (Option C)</span>
6 0
4 years ago
Read 2 more answers
Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you
Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}

With \pi, we should expect \cos x. If \tan x=\dfrac8{15}, then

\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}

Also,

\pi

so we should expect \tan\dfrac x2 and

\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4

You seem to be taking

\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}

but this is not an identity.

###

8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

\dfrac{\sqrt6-\sqrt2}4

###

9. Use the identity from (7). \dfrac{7\pi}{12} lies in the second quadrant, so

\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3

###

12.

\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}

=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}

=\dfrac{\cos x}{\sin x}

=\dfrac{\csc x}{\sec x}

###

13.

\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}

=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}

=\dfrac1{\cos x}

=\sec x

###

14.

\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)

=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}

=2

###

15.

\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}

=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}

=\cos2\theta

3 0
3 years ago
Jim ran 3/4 of a mile everyday after school for two weeks. how many miles did he run each day?
bagirrra123 [75]
3/4 X 7/1=5 1/4 Then, 5 1/4+5 1/4=10 1/2. Answer 10 1/2
7 0
3 years ago
Read 2 more answers
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