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4 years ago

Which pairs of angles in the figure below are vertical angles?

Mathematics

2 answers:

SIZIF [17.4K]4 years ago

7 0

Answer:

B and C

Step-by-step explanation:

Vertical angles are the angles opposite of each other.

amm18124 years ago

5 0

Answer: ISN and TSW or TSN and ISW same thing so b and c

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Solve the following system of equations using the substitution method.

BartSMP [9]

Answer:

Step-by-step explanation:

First label the equations.

6x +3y= 18 -----(1)

y= -2x +5 -----(2)

subst. (2) into (1):

6x +3(-2x +5)= 18

6x +3(-2x) +3(5)= 18 (expand)

6x -6x +15= 18

15= 18 (simplify)

(reject)

There is no solution.

<u>Further explanation</u><u>:</u>

If I rewrite the first equation into the form of y=mx+c, I would get

6x +3y=18

3y= 18 -6x (bring y term to one side)

y= 6- 2x (÷3 throughout)

y= -2x +6

Notice that the gradients of both lines are equal ( -2). This implies that they are parallel to each other and will never meet. Hence there are no solutions.

3 0

3 years ago

Which inverse operation would be used to verify the following equation? 102 ÷ 3 = 34

IRISSAK [1]

The answer would be 102/34 = 3

7 0

3 years ago

Read 2 more answers

Solve for x. X^2 = 1/4. Look at the picture for answer choices. Don't answer if you dont know.

allsm [11]

Answer:

x is equal to ± 1/2

Step-by-step explanation:

To find x, we simply need to take the square root of both sides:

x² = 1/4

√(x²) = √(1/4)

x = √(1/4)

x = ± 1/2

5 0

3 years ago

Q1 A ball is thrown upwards with some initial speed. It goes up to a height of 19.6m and then returns. Find (a) The initial spee

lubasha [3.4K]

Answer:

(a) 19.6 ms⁻¹

(b) 2 s

(d) 4 s

Step-by-step explanation:

<u>Constant Acceleration Equations (SUVAT)</u>

$\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}$

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Acceleration due to gravity = 9.8 ms⁻².

When the ball reaches its maximum height, its velocity will momentarily be zero.

<u>Given values</u> (taking up as positive):

$s=19.6 \quad v=0 \quad a=-9.8$

$\begin{aligned}\textsf{Using} \quad v^2&=u^2+2as\\\\\textsf{Substitute the given values:}\\0^2&=u^2+2(-9.8)(19.6)\\0&=u^2-384.16\\u^2&=384.16\\u&=\sqrt{384.16}\\\implies u&=19.6\; \sf ms^{-1}\end{aligned}$

Therefore, the initial speed is 19.6 ms⁻¹.

Using the same values as for part (a):

$\begin{aligned}\textsf{Using} \quad s&=vt-\dfrac{1}{2}at^2\\\\\textsf{Substitute the given values:}\\19.6&=0(t)-\dfrac{1}{2}(-9.8)t^2\\19.6&=4.9t^2\\t^2&=\dfrac{19.6}{4.9}\\t^2&=4\\t&=\sqrt{4}\\\implies t&=2\; \sf s\end{aligned}$

Therefore, the time taken to reach the highest point is 2 seconds.

As the ball reaches its maximum height at 2 seconds, one second before this time is 1 s.

<u>Given values</u> (taking up as positive):

$u=19.6 \quad a=-9.8 \quad t=1$

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=19.6+(-9.8)(1)\\v&=19.6-9.8\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

The velocity of the ball one second before it reaches its maximum height is the <u>same</u> as the velocity one second after.

<u>Proof</u>

When the ball reaches its maximum height, its velocity is zero.

Therefore, the values for the downwards journey (from when it reaches its maximum height):

$u=0 \quad a=9.8 \quad t=1$

(acceleration is now positive as we are taking ↓ as positive).

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=0+9.8(1)\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

Therefore, the velocity of the ball one second before <u>and</u> one second after it reaches the maximum height is 9.8 ms⁻¹.

From part (a) we know that the time taken to reach the highest point is 2 seconds. Therefore, the time taken by the ball to travel from the highest point to its original position will also be 2 seconds.

Therefore, the total time taken by the ball to return to its original position after it is thrown upwards is 4 seconds.

4 0

1 year ago

HELP PLZZ!! ill give brainlist

Neko [114]

Answer:

Step-by-step explanation:

The spinner should land on whatever color 25% of the time.

25% of 20 = 25/100 * 20 = 500/100 = 5

So the correct answer is the number of times the spinner lands on a color 5 times. Only white does that. All the other colors either go above 5 or below it.

7 0

3 years ago

Read 2 more answers