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2 years ago

Find the quotient: -40 ÷ 5

Mathematics

2 answers:

Hoochie [10]2 years ago

5 0

The answer for this would be negative eight

GalinKa [24]2 years ago

5 0

-8

Hope this helps

Brainliest would be nice

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Which of the following is not a proportion 1/4=3/12 4/10=12/20 2/4=3/6 8/6=20/15

Lera25 [3.4K]

The answer is 4/10 = 12/20.

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3 years ago

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What is the image of (-1,-5) after a dilation by a scale factor of 5 centered at the origin?

Travka [436]

A dilation by a scale factor of 5 centered at the origin simply multplies by 5 the coordinates of every point. So, the new coordinates are (-5, -25)

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3 years ago

Find Sin (BAC+30) 80POINTS

Ann [662]

Answer:

$\dfrac{2}{5}\sqrt{3}+\dfrac{3}{10}$

Step-by-step explanation:

Trigonometric Identities

$\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B$

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

$\cos (\angle BAC)=\dfrac{12}{20}$

$\sin (\angle BAC)=\dfrac{16}{20}$

Angles

$\sin (30^{\circ})=\dfrac{1}{2}$

$\cos (30^{\circ})=\dfrac{\sqrt{3}}{2}$

Therefore, using the trig identities and ratios:

$\begin{aligned}\sin (\angle BAC + 30^{\circ}) & = \sin (\angle BAC) \cos (30^{\circ})+\cos (\angle BAC) \sin (30^{\circ})\\\\& = \dfrac{16}{20} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{12}{20} \cdot \dfrac{1}{2}\\\\& = \dfrac{16}{40}\sqrt{3}+\dfrac{12}{40}\\\\& = \dfrac{2}{5}\sqrt{3}+\dfrac{3}{10}\end{aligned}$

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1 year ago

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Becky and tommy are asked to solve -4x-25=8x+15 identity where one of them made an error

ololo11 [35]

I feel sorry for Becky and Tommy. The one who didn't come up with x=-10/3 made an error.

first bring 8x to the left and -25 to the right and simplify:

-4x-8x = 25+15 => -12x = 40

x = -40/12 = -10/3

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3 years ago

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I need help with questions #7 and #8 plz

katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

$c^2 = a^2 + b^2 - 2ab \cos C$

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

We use the law of cosines to find C.

$18^2 = 12^2 + 16^2 - 2(12)(16) \cos C$