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Alenkasestr [34]
3 years ago
6

here are some numbers 8 11 13 18 25 37 , 8 13 18 is an arithmetic progression and the rule is to add 5 use 3 of the numbers to m

ake a different arithmetic progression.
Mathematics
1 answer:
chubhunter [2.5K]3 years ago
3 0

Answer:

11, 18, 25

Step-by-step explanation:

add 7

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ser-zykov [4K]

3^{x + 2} - 3^{x} = 3^{x}* 3^{2} - 3^{x} = 9(3^{x}) - 1(3^{x}) = 8(3^{x})

3^{x + 2} - 3^{x} = \frac{8}{9}

⇒ 8(3^{x}) = \frac{8}{9}

⇒ 3^{x} = \frac{1}{9}   <em>multiplied both sides by 8</em>

⇒3^{x} = \frac{1}{3^{2} }

⇒ 3^{x} = 3⁻²

⇒ x = -2

Answer: x = -2

6 0
3 years ago
For what values of does the equation (2 + 1)^2 + 2 = 10 − 6 have two real and equal roots?
Tatiana [17]

Answer:

(2 + 1)^2 + 2 = 10 - 6  

Falso

Step-by-step explanation:

109 / 5000  

El lado izquierdo

11

 no es igual al lado derecho

4

, lo que significa que la declaración dada es falsa.

Falso

3 0
3 years ago
A tree grows vertically on a hillside. The hill is at a 16° angle to the horizontal. The tree casts an 18-meter shadow up the hi
elixir [45]
The height of the tree is the sum of the part below the line parallel to the horizontal and the part above the line parallel to the horizontal.

Height of the part below the line parallel to the horizontal = 18 sin16° = 4.96 meters
Horizontal distance of the tip of the of the shadow from the tree = 18 cos16° = 17.30 meters
Height of the part above the line parallel to the horizontal = 17.3 tan68° = 42.83 meters

Height of the tree = 4.96 + 42.83 = 47.79 meters
8 0
3 years ago
If sinθ = -1/2 and θ is in Quadrant III, then tanθ = _____.
Gnom [1K]

Answer:  \tan \theta=\dfrac{1}{\sqrt3}.

Step-by-step explanation:  Given that

\sin\theta=-\dfrac{1}{2} and \theta lies in Quadrant III.

We are to find the value of \tan \theta.

We will be using the following trigonometric identities:

(i)~sin^2\theta+\cos^2\theta=1,\\\\(ii)~\dfrac{\sin\theta}{\cos{\theta}}=\tan \theta.

We have

\tan\theta\\\\\\=\dfrac{\sin\theta}{\cos\theta}\\\\\\=\dfrac{\sin\theta}{\pm\sqrt{1-\sin^2\theta}}\\\\\\=\pm\dfrac{-\frac{1}{2}}{\sqrt{1-\left(\frac{1}{2}\right)^2}}\\\\\\=\pm\dfrac{\frac{1}{2}}{\sqrt{1-\frac{1}{4}}}\\\\\\=\pm\dfrac{\frac{1}{2}}{\frac{\sqrt3}{2}}\\\\\\=\pm\dfrac{1}{\sqrt3}.

Since \theta lies in Quadrant III, so tangent will be positive.

Thus,

\tan \theta=\dfrac{1}{\sqrt3}.

8 0
4 years ago
Read 2 more answers
Sun-Yi estimated 270+146and got 300.is her estimate reasonable explain
Feliz [49]
No. Round 146 to 150 and add both 270 and 150. It's 420. So 400 would be more reasonable. 300 is not reasonable because it ain't close.
3 0
4 years ago
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