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IRISSAK [1]
2 years ago
5

I need help in solving a math problem.

Mathematics
1 answer:
Fittoniya [83]2 years ago
7 0

Negative value has least value.


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*20 pts*
Leno4ka [110]

a.      This is the geometric sequence: a1 = 1.5 m = 150 cm, r = 0.74

The formula now is an = a1 x r ^ n-1

 

Just substitute a1 = 150 cm and r = 0.74

an = 150 x (0.74) ^ n – 1

 

b.      The second part is:

A6 = 150 x 0.74 ^ 6-1

= 150 x 0.74 ^5

= 150 x 0.221906624

= 33.28509936 cm

4 0
3 years ago
Read 2 more answers
Help me... I don't really understand and need an explanation
ehidna [41]

Answer:

x=17

Step-by-step explanation:

This should be for a "kite": AD=DC, AB=BC and ∠A=∠C

8x-27 = 3x+58

5x = 85

x = 17

3 0
2 years ago
What is h^2+6h-27 ????????????​
nydimaria [60]

Answer:

Sorry it's acually (h+3)(h-9)

4 0
3 years ago
Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab
Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

<u>Left part:</u>

\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}

<u>Right part:</u>

\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}

Since simplified left and right parts are the same, then the equality is true.

3 0
3 years ago
Cannon balls for an antique cannon are stacked next to it in four layers to form a square pyramid. How many cannon balls are in
labwork [276]
The amount of balls in a given layer can be found by squaring the layer it is in

4 layers
top layer (1st) is 1
2nd is 4
3rd layer is 9
4th is 16

add them up
1+4+9+16=30

30 in that pyramid
7 0
3 years ago
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