All categories

4 years ago

A triangle has side lengths measuring 3x cm, 7x cm, and h cm. Which expression describes the possible values of h, in cm?

Mathematics

2 answers:

GREYUIT [131]4 years ago

8 0

Hi,

Sum of two sides of triangle always greater than third side, so the possible values of h is:

7x-3x < h < 7x+3x

4x < h < 10x

I hope to help you☻♥

nata0808 [166]4 years ago

3 0

Answer:

Option A.

Step-by-step explanation:

A triangle has 3 sides measuring 3x cm, 7x cm and h cm.

We have to determine the expression which describes the possible value of 'h'

Since a triangle is possible if " Sum of two sides of a triangle is greater than the measure of third side."

so h + 3x > 7x

h > (7x - 3x)

h > 4x

or 3x + 7x > h

10x > h

So by combining both inequalities

4x < h < 10x will be the answer.

Option A is the answer.

You might be interested in

Which COULD be the area of this parallelogram?

BaLLatris [955]

Answer:

1650 in

Step-by-step explanation:

base times height

50 times 33 = 1650 in

7 0

3 years ago

Read 2 more answers

-1 - (-3)= Please answer this if you can please explane it or show the work?

Arisa [49]

Answer:

Step-by-step explanation:

youre subtracting a negative, which makes it positive. anytime you subtract a negative, its just addition :)

7 0

3 years ago

Read 2 more answers

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

A particular metal has a density of 25g/cm cubed. Calculate the mass of 178cm cubed of this metal.

mel-nik [20]

Mass = density x volume

Mass = 178 x 25 = 4,450 grams

6 0

3 years ago

The product of 3.3 and a number C

ladessa [460]

The " product " is the result of multiplication.
3.3C is ur answer

8 0

3 years ago