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Mashutka [201]
3 years ago
5

Help pls !! this is my trig test btw ​

Mathematics
1 answer:
gulaghasi [49]3 years ago
4 0
<h3><u>Required Answer:-</u></h3>

This is an right angle ∆ and the side lengths containing a right angle are 54 cm and 85 cm.

By Pythagoras theoram,

{a}^{2}  +  {b}^{2}  =  {c}^{2}

where a is the perpendicular, b is the base and c is the hypotenuse.

Plugging the values,

{54}^{2}  +  {85}^{2}  =  {h}^{2}

Then,

h =  \sqrt{ {54}^{2}  +  {85}^{2} }

h = 100.70 cm(approx.)

<h3><u>Hence:-</u></h3>

The x of the right angled ∆ = <u>100.</u><u>7</u><u>0</u><u> </u><u>cm</u>

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If (a^3+27)=(a+3)(a^2+ma+9) then m equals
Ronch [10]

Answer:

m = - 3

Step-by-step explanation:

a³ + 27 ← is a sum of cubes and factors in general as

a³ + b³ = (a + b)(a² - ab + b²), thus

a³ + 27

= a³ + 3³

= (a + 3)(a² - 3a + 9)

comparing a² - 3a + 9 to a² + ma + 9, then

m = - 3

7 0
3 years ago
Hello, precalc, need help on finding csc
Ainat [17]

Recall the double angle identity for cosine:

\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - 2 \sin^2(x)

It follows that

\sin^2(x) = \dfrac{1 - \cos(2x)}2 \implies \sin(x) = \pm \sqrt{\dfrac{1-\cos(2x)}2} \implies \csc(x) = \pm \sqrt{\dfrac2{1-\cos(2x)}}

Since 0° < 22° < 90°, we know that sin(22°) must be positive, so csc(22°) is also positive. Let x = 22°; then the closest answer would be C,

\csc(22^\circ) = \sqrt{\dfrac2{1-\cos(44^\circ)}} = \sqrt{\dfrac2{1-\frac5{13}}} = \dfrac{\sqrt{13}}2

but the problem is that none of these claims are true; cot(32°) ≠ 4/3, cos(44°) ≠ 5/13, and csc(22°) ≠ √13/2...

3 0
3 years ago
Solve the equation by completing the square. Round the square. Round to the nearest hundredth if necessary. x2 - 3x -12 =0
dmitriy555 [2]
A:5.27,-2.27 should be your answer
5 0
3 years ago
Select the correct answer. The radius of a nitrogen atom is 5. 6 × 10-11 meters, and the radius of a beryllium atom is 1. 12 × 1
Grace [21]

A negative exponent means the number of the exponent divide by the number. The the radius of beryllium atom is larger than the radius of a nitrogen atom by two times.

Given information-

The radius of a nitrogen atom is 5.6\times10^{-11} meters.

The radius of a beryllium atom is 1.12 \times10^{-10} meters

<h3>Negative exponents</h3>

A negative exponent means the number of the exponent divide by the number.

As the radius of the nitrogen atom is 5.6\times10^{-11} meters. Convert it in the power of negative exponent of 10.

Let the radius of the nitrogen is r mm. Thus,

r=5.6\times 10^{-11}\\&#10;r=5.6\times 10^{-10-1}\\&#10;r=5.6\times 10^{-10}\times10^{-1}\\&#10;r=\dfrac{5.6\times 10^{-10}}{10} \\&#10;r=0.56\times 10^{-10}

Now the base of the exponents of both the atoms are same. As the radius of the nitrogen atom is 0.56\times 10^{-10} meters and the radius of the beryllium atom is 1.12 \times10^{-10}. Thus the radius of beryllium atom is larger.

To calculate how many times, the radius of beryllium atom should be divided by the radius of a nitrogen atom. Thus,

=\dfrac{1.12\times10^{-10}}{0.56\times10^{-10}} \\&#10;=2

Thus the the radius of beryllium atom is larger than the radius of a nitrogen atom by two times.

Learn more about the negative exponents here;

https://brainly.in/question/1221452

7 0
3 years ago
HELP PLS ITS OVER DUE PLEASE MATCH THESE! please
Radda [10]
I’ve attached a photo with the definition and my answers
Hope this helps!

6 0
3 years ago
Read 2 more answers
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