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Hitman42 [59]
3 years ago
5

Clock wise angle from minutes hand to hour hand at 7 o'clock is what (a)240° (b)230° (c)220° (d)210° (e)200°

Mathematics
2 answers:
AlladinOne [14]3 years ago
7 0

Answer:

Option D) Angle at 7 o'clock = 210 degrees

Step-by-step explanation:

The whole clock = 360 degrees

Every 1 hour = 360/12

Every 1 hour = 30 degrees

Angle at 7 o'clock = 30*7

Angle at 7 o'clock = 210 degrees

lord [1]3 years ago
6 0

Answer:

d

Step-by-step explanation:

At 7 o' clock the minute hand is at 12 and the hour hand at 7

There is 360° in a complete revolution, thus between 7 and 12

angle = \frac{7}{12} × 360° = 7 × 30° = 210° → d

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so that 15 fits into 20 at 3/4

1- 3/4 = 1/4

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Step-by-step explanation:

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2 years ago
What are the slope and y-intercept for the graph of y – 3x = -1?
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Answer:

Slope (m) = 3, y-intercept (b) = -1

Step-by-step explanation:

First, we need to write the equation in y = mx + b.

To do this, add 3x to both sides.

y = 3x - 1 is your result.

Now, we have everything we need to do determine our slope and y-intercept in just this equation alone.

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2 years ago
NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 11z​
Roman55 [17]

Answer:

(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}

(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}

Step-by-step explanation:

Given system of equations:

\begin{cases}y=x^2-9\\y=4x-4\end{cases}

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

\begin{aligned}x^2-9&=4x-4\\x^2-4x-9&=-4\\x^2-4x-5&=0\end{aligned}

Factor the quadratic:

\begin{aligned}x^2-4x-5&=0\\x^2-5x+x-5&=0\\x(x-5)+1(x-5)&=0\\(x+1)(x-5)&=0\end{aligned}

Apply the <u>zero-product property</u> and solve for x:

\implies x+1=0 \implies x=-1

\implies x-5=0 \implies x=5

Substitute the found values of x into the <u>second equation</u> and solve for y:

\begin{aligned}x=-1 \implies y&=4(-1)-4\\y&=-4-4\\y&=-8\end{aligned}

\begin{aligned}x=5 \implies y&=4(5)-4\\y&=20-4\\y&=16\end{aligned}

Therefore, the solutions are:

(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}

(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}

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