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3 years ago

What is the value of p?

Mathematics

2 answers:

kenny6666 [7]3 years ago

4 0

Answer:

Step-by-step explanation:

All angles of a triangle add up to 180 degrees, so you can add the other two angles, then subtract by 180.

Nikitich [7]3 years ago

3 0

35 would be the answer

You might be interested in

Please answer someone!!!

Step2247 [10]

Answer:

True

Step-by-step explanation:

$\sqrt{(xy)^3}$ =
$((xy)^{1/2})^3$ =
$(xy)^{{1/2}*3}$ =
$(xy)^{3/2}$

The answer is TRUE

8 0

3 years ago

A man who can ride a bicycle 50 miles in 2 hrs and 25 minutes has what average speed in mph?

patriot [66]

145/50=2.9
2.9mph on average

5 0

3 years ago

Need help asappppppp

PilotLPTM [1.2K]

Given:

The right triangular prism.

Height of prism = 28 in.

Hypotenuse of base = 25 in.

leg of base = 24 in.

To find:

The lateral surface area of the prism.

Solution:

Pythagoras theorem:

$Hypotenuse^2=Perpendicular^2+Base^2$

Using Pythagoras theorem in the base triangle, we get

$25^2=24^2+Base^2$

$625-576=Base^2$

$\sqrt{49}=Base$

$7=Base$

The perimeter of the triangular base is:

$P=7+25+24$

$P=56$

Lateral area of a triangular prism is:

$A=Ph$

Where, P is the perimeter of the triangular base and h is the height of the prism.

Putting $P=56,h=28$ in the above formula, we get

$A=56(28)$

$A=1568$

Therefore, the lateral area of the prism is 1568 in².

8 0

3 years ago

Plz help and ty very much

IgorC [24]

Answer:

True

Step-by-step explanation:

For a compound inequality using the word "or" to be true, all you need is for one of the parts to be true.

-6 < 7 is true

-2 <= -2 is also true

Answer: True

4 0

3 years ago

The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of the bridge is 1210 m long and

brilliants [131]

Answer:

The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

Step-by-step explanation:

The equation of the parabola is:

$y=0.00035x^{2}$

Compute the first order derivative of y as follows:

$y=0.00035x^{2}$

$\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^{2}]$

$=2\cdot 0.00035x\\\\=0.0007x$

Now, it is provided that |x | ≤ 605.

⇒ -605 ≤ x ≤ 605

Compute the arc length as follows:

$\text{Arc Length}=\int\limits^{x}_{-x} {1+(\frac{\text{dy}}{\text{dx}})^{2}} \, dx$

$=\int\limits^{605}_{-605} {\sqrt{1+(0.0007x)^{2}}} \, dx \\\\={\displaystyle\int\limits^{605}_{-605}}\sqrt{\dfrac{49x^2}{100000000}+1}\,\mathrm{d}x\\\\={\dfrac{1}{10000}}}{\displaystyle\int\limits^{605}_{-605}}\sqrt{49x^2+100000000}\,\mathrm{d}x\\\\$

Now, let

$x=\dfrac{10000\tan\left(u\right)}{7}\\\\\Rightarrow u=\arctan\left(\dfrac{7x}{10000}\right)\\\\\Rightarrow \mathrm{d}x=\dfrac{10000\sec^2\left(u\right)}{7}\,\mathrm{d}u$

$\int dx={\displaystyle\int\limits}\dfrac{10000\sec^2\left(u\right)\sqrt{100000000\tan^2\left(u\right)+100000000}}{7}\,\mathrm{d}u$

$={\dfrac{100000000}{7}}}{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u\\\\=\dfrac{50000000\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{7}+\dfrac{50000000\sec\left(u\right)\tan\left(u\right)}{7}\\\\=\dfrac{50000000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+5000x\sqrt{\dfrac{49x^2}{100000000}+1}$

Plug in the solved integrals in Arc Length and solve as follows:

$\text{Arc Length}=\dfrac{5000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+\dfrac{x\sqrt{\frac{49x^2}{100000000}+1}}{2}|_{limits^{605}_{-605}}\\\\$

$=1245.253707795227\\\\\approx 1245.25$

Thus, the approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

7 0

3 years ago