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4 years ago

I need the answer plz

Mathematics

1 answer:

lakkis [162]4 years ago

5 0

Answer:

GH = 7 $\sqrt{2}$

Step-by-step explanation:

Using the sine ratio in the right triangle and the exact value

sin45° = $\frac{1}{\sqrt{2} }$

sin45° = $\frac{opposite}{hypotenuse}$ = $\frac{HI}{GH}$ = $\frac{7}{GH}$ = $\frac{1}{\sqrt{2} }$ ( cross- multiply )

GH = 7 $\sqrt{2}$

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A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line

Nataliya [291]

Answer:

Volume is $2000\pi\ m^{3}$

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

$c = \frac{1}{5}$

Therefore,

$h(y) = \frac{1}{5}y + 5$

Now, the Volume of the pool is given by:

$V = \int h(y)dA$

where

A = $r\theta$

$A = rdr\ d\theta$

Thus

$V = \int (\frac{1}{5}y + 5)dA$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta$

$V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta$

$V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}$

$V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}$

7 0

3 years ago

Which exponential function is represented by the values in the table? (-2,16) (-1,8) (0,4) (1,2) (2,1)

mote1985 [20]

4(r)^x=y
4(r)^(1)=2
r=1/2
y=4*(1/2)^x

7 0

4 years ago

Given that Cosecant (t) = negative StartFraction 13 Over 5 EndFraction for Pi less-than t less-than StartFraction 3 pi Over 2 En

Sergio [31]

Answer:

$\bold{cot(t) =\dfrac{12}{5}}$

Step-by-step explanation:

Given that:

$Cosec (t) = -\frac{13}5$

for $\pi < t < \frac{3 \pi}2$

That means, angle $t$ is in the 3rd quadrant.

To find:

Value of cot(t)

Solution:

First of all, let us recall what trigonometric ratios are positive and what trigonometric ratios are negative in 3rd quadrant.

In 3rd quadrant, tangent and cotangent are positive.

All other trigonometric ratios are negative.

Let us have a look at the following identity:

$cosec^2\theta -cot^2\theta =1$

here, $\theta =t$

So, $cosec^2t-cot^2t=1$

$\Rightarrow (-\dfrac{13}{5})^2-cot^2t=1\\\Rightarrow (\dfrac{169}{25})-cot^2t=1\\\Rightarrow \dfrac{169}{25}-1=cot^2t\\\Rightarrow \dfrac{169-25}{25}=cot^2t\\\Rightarrow \dfrac{144}{25}=cot^2t\\\Rightarrow cot(t)=\pm\sqrt{\dfrac{144}{25}}\\\Rightarrow cot(t)=\pm\dfrac{12}{5}$

But, angle $t$ is in 3rd quadrant, so value of

$\bold{cot(t) =\dfrac{12}{5}}$

4 0

3 years ago

It took Aiden 6 hours to plant 8 trees. How many hours would it take for him to plant 20 trees? 24, 18, 23, 15

Pepsi [2]

<h3>Answer: 15 hours (choice D)</h3>

================================================

Work Shown:

(6 hours)/(8 trees) = (x hours)/(20 trees)

6/8 = x/20

6*20 = 8*x ... cross multiply

120 = 8x

8x = 120

x = 120/8

x = 15

It takes him 15 hours to plant 20 trees.

7 0

3 years ago

Find the value for x to the third decimal place: In x=5

mixas84 [53]

$\ln x=5\implies x=e^5\approx0.6738\cdot10^{-2}$ .

Hope this helps.

8 0

3 years ago