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3 years ago

I need help with math

Mathematics

1 answer:

aksik [14]3 years ago

8 0

Well what is the problem ?

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Will give brainliest if done properly

k0ka [10]

Answer:

$\sf \boxed{ \sf y - 3 = \dfrac{3}{2} (x - 6)}$

Explanation:

$\sf Given \ equation: y = \dfrac{3}{2} x - 5$

Comparing it with slope intercept form "y = mx + b" where 'm' is slope and 'b' is y-intercept.

Here slope: 3/2 and y-intercept: -5

Parallel slope has the same tangent slope.

Pass through point (x, y) = (6, 3)

Equation:

$\sf y - y_1 = m(x - x_1)$

$\rightarrow \sf y - 3 = \dfrac{3}{2} (x - 6)$

6 0

1 year ago

Find the slope of the line containing the points (5, -1) and (-8, -4).

UkoKoshka [18]

I think it might be D

7 0

2 years ago

Use the circle to represent the following situation and show how you came up with the answer.

frozen [14]

Answer:

look below bestie

Step-by-step explanation:

ok so i dont know most of it (sryy) but the area of the entire pizza is 535.84in², the radius of the pizza is about 13.05 and yea. thats all i got

3 0

2 years ago

Please help, and i will mark you BRAINLIEST <br><br> QUESTION:<br> Solve: (-8) x (-4) = ________

aleksandr82 [10.1K]

Answer:

Step-by-step explanation:

Both are negative so it come out to positive and then 8*4 is 32

5 0

3 years ago

1)a chord of length 18cm midway the radius of a circle. calculate the radius of the circle correct to 1d.p. 2)if two parallel ch

kykrilka [37]

Part 1:

Given that the length of the chord is 18 cm and the chord is midway the radius of the circle.

Thus, half the angle formed by the chord at the centre of the circle is given by:

$\cos\theta=\frac{\left( \frac{1}{2} r\right)}{r}= \frac{1}{2} \\ \\ \Rightarrow\theta=\cos^{-1}\left( \frac{1}{2} \right)=60^o$

Now,

$\sin60^o= \frac{9}{r} \\ \\ \Rightarrow r= \frac{9}{\sin60^o} =10.392$

Therefore, the radius of the circle is 10.4 cm to 1 d.p.

Part 2I:

Given that the radius of the circle is 10 cm and the length of chord AB is 8 cm. Thus, half the length of the chord is 4cm. Let the distance of the mid-point O to /AB/ be x and half the angle formed by the chord at the centre of the circle be θ, then

$\sin\theta= \frac{4}{10} = \frac{2}{5} \\ \\ \theta=\sin^{-1}\left( \frac{2}{5} \right)=23.6^o$

Now,

$\cos23.6^o= \frac{x}{10} \\ \\ \Rightarrow x=10\cos23.6^o=9.165\approx9.2cm$

Part 2II:

Given that the radius of the circle is 10cm and the angle distended is 80 degrees. Let half the length of chord CD be y, then:

$\sin40^o= \frac{y}{10} \\ \\ \\ \Rightarrow y=10\sin40^o=6.428$

Thus, the length of chord CD = 2(6.428) = 12.856 which is approximately 12.9 cm.

3 0

3 years ago