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Taya2010 [7]
3 years ago
5

Factorise 12y+16y need help with maths homework

Mathematics
2 answers:
MissTica3 years ago
5 0
4y(3 + 4), because they both have 4y in common, and if you multiply 4y by 3 you get 12y, and multiplied by 4 you get 16y. Hope this helps!
poizon [28]3 years ago
3 0
The answer is    28y hope this helps

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Ivan

the height:

h = \frac{P}{mg}

(with g being the gravitational acceleration)

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3 years ago
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8 0
3 years ago
Please solve the following questions!!
alexandr402 [8]

Answer:

3. The missing angle is 56°

4. x = 7

Step-by-step explanation:

3.

We know sum of 3 angles in a triangle is 180°.

Looking at the top triangle, we can figure out the third angle. Let third angle be x:

85 + 35 + x = 180

120 + x = 180

x = 180 - 120

x = 60

<u>The angle "x" and the angle that is missing from the "bottom" triangle in the figure, are vertical angles, and hence, are EQUAL.</u>

So the bottom triangle now has 2 angles, 60 and 64 (given). Let the third angle be y(the one with a question mark). So we can write:

60 + 64 + y = 180

124 + y = 180

y = 180 - 124

y= 56

This is the missing angle.

4.

10x - 5 AND 8x + 9 are vertical angles. They ARE EQUAL.

Thus we can write the equation:

(10x-5) =  (8x+9)

10x-8x=9+5

2x=14

x=14/2

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5 0
3 years ago
The equation of a circle is (x - 3)^2 + (y + 2)^2 = 25. The point (8, -2) is on the circle.
Jlenok [28]
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8 0
3 years ago
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Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta

Simplify. The denominator uses the difference of two squares pattern:

\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta

Split into two separate fractions:

\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta

Rewrite the two fractions:

\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=}  \sec^2\theta - \sec\theta\tan\theta

Hence verified.

8 0
3 years ago
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