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3 years ago

Please help I have 3 questions.

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

7 0

Question 1:
The two triangles that are congruent by ASA are PQR and VXW (It should be).
ASA congruence are two angles with the side between the two angles being congruent.
Since I don't see any of the answer choices that match this, the second choice is most likely the answer. There might be typo.

Question 2:
One of the correct triangles if flipped to make it harder.
The answers are CAB and FED (since that the only choice with "matching" letters), or the third choice.

Question 3:
In order to prove this triangle is congruent by ASA, T must be congruent to S.
We know that Q is congruent to Q for both triangles.
The first choice is the answer.

Have an awesome day! :)

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Y-intercept = (0, 2)<br> slope = -3/7 what is the equation?

Marianna [84]

Answer:

y = -(3/7)x + 2

Step-by-step explanation:

(see attached)

recall that the slope-intercept form of a linear equation is

y = mx + b

where m = slope = given as -(3/7)

and b = y-intercept = 2

substituting these values into the eqation:

y = mx + b

y = -(3/7)x + 2

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3 years ago

Find the surface area of the rectangular prism.

finlep [7]

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3 years ago

What property is in (6+3)+21=6+(3+21)

Assoli18 [71]

This is called the Associative Property of Addition.

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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.

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1 year ago

Ted has $55. 00 in his pocket. Which purchase will he be able to pay for with cash? two books for $29. 95 each headphones for $2

kotykmax [81]

If Ted has $55. 00 in his pocket then out of two books for $29. 95 each headphones for $29. 99 and a portable speaker set for $30. 95 an MP3 player for $35. 99 and a $20. 00 gift card for downloading music three books for $16. 99 each he can purchase three books for $16. 99 each.

<h3>How to calculate total money spent ?</h3>

To calculate total money spent we have to add all the money we spent on all things.

Given that Ted has $55. 00 in his pocket.

(i)

two books for $29. 95 each

Money needed for 2 books = $2\times 29.95=59.9$

So Ted is not able to purchase 2 books

(ii) headphones for $29. 99 and a portable speaker set for $30. 95

Money needed for headphones and portable speaker = 29.99+30.95=60.94

So Ted is not able to purchase headphones and portable speaker

(iii) an MP3 player for $35. 99 and a $20. 00 gift card for downloading music

Money needed for MP3 player and gift card= 35.99+20.00=55.99

So Ted is not able to purchase MP3 player and gift card

(iv) three books for $16. 99 each.

Money needed for 3 books = $3\times16.99$ =50.97

So ted is able to purchase these three books .

To learn more about money spent visit : brainly.com/question/21625005

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2 years ago