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3 years ago

Select all the true sentences!!!

Mathematics

1 answer:

iragen [17]3 years ago

6 0

Answer:

the second one and third one

Step-by-step explanation:

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What is the ratio of the area of abc to the area of triangle qst..?

Ivahew [28]

The ration of a triangle is the same as a is to b as b is to c and c is to a.

6 0

3 years ago

The polynomial −16t^2+550 gives the height of a ball t seconds after it is dropped from a 550-foot-tall building. Find the heigh

zhenek [66]

Answer:

486 feet

Step-by-step explanation:

h = -16t² + 550

h = -16(2)² + 550

h = 486

7 0

3 years ago

If x= 8 x 9 x 27 then what is 3 square root of x

forsale [732]

Answer: 132.27

Step-by-step explanation:

Multiply 8 × 9 × 27 to get the value of x

= 1944

To get 3 square root of 1944, find the square root of 1944 = 44.09

3 square root of 1944 = 3 x 44.09 = 132.27

6 0

3 years ago

Estimate √97.5 / 1.96

evablogger [386]

Answer: 5

Step-by-step explanation:

1. You have the following expression given in the problem above:

$\frac{\sqrt{97.5}}{1.96}$

2. You can estimate the result by rounding the numerator and the denominator.

3. As you can see, you can round up 97.5 to 100.

4. Then, you can round up 1.96 to 2.

5. Therefore, you have:

$\frac{\sqrt{100}}{2}$

$\frac{10}{2}=5$

6. Therefore, the result is 5.

3 0

3 years ago

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

2 years ago