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3 years ago

Please help! Picture is included! This is due very soon and I need help! Number 11. (10 points)

Mathematics

1 answer:

AlekseyPX3 years ago

8 0

Y+y+17+111+90=360
2y+128+90=360
2y=360-218
2y=142
y=142/2
y=71
thus, the angles are 71, 90, 111, 88.

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Rami designed a small pond for a restaurant. The diagram below shows the measurements of the pond. How many cubic feet of water

Naddik [55]

Answer:

426

Step-by-step explanation:

6 0

2 years ago

For questions 1 – 6, find the area of the circle to the nearest hundredth.<br> Any help appriciated!

Ganezh [65]

Step-by-step explanation:

The area of a circle is given by :

A = πr²

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(4) Diameter = 22 cm

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$A=\pi \times (11)^2\\\\A=380.13\ cm^2$

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$A=\pi \times (2)^2\\\\A=12.56\ cm^2$

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4 0

2 years ago

A chord is 10cm from the centre of a circle of radius 15cm. Find the length of the chord.

almond37 [142]

Answer: 21.44 cm

Step-by-step explanation:

Given

Radius r=15 cm

distance of chord from the center is 10 cm

From the figure, the length of the chord is AC

Applying Pythagoras

$\Rightarrow AO^2=OB^2+AB^2\\\Rightarrow AB^2=15^2-10^2\\\Rightarrow AB^2=225-100=115\\\Rightarrow AB=\sqrt{115}=10.72\ cm$

The length of the chord is $AC=2AB=21.44\ cm$

5 0

3 years ago

Choose the fraction that has not been reduced to simplest form.

Thepotemich [5.8K]

first one since it’s 1/16 simplest form or 11/17 cuz it’s obv

8 0

3 years ago

Read 2 more answers

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

3 0

3 years ago