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3 years ago

HELP! Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Mathematics

2 answers:

OverLord2011 [107]3 years ago

7 0

Answer:

f(x) = 3.2^n-1

f(x) = 2x4 ^n-1

f(x) = 2x6^n-1

Step-by-step explanation:

f(1) = 3x2^(1-1)

= 3x1

= 3

f(2)= 3x2^n-1

= 3x2 ^(2-1)

= 3x2

= 6

f(3)= 3x2^(3-1)

= 3x2^2

= 3x4

= 12

f(1) = 2x4^(1-1)

= 2x1

= 2

f(2) = 2x4^(2-1)

= 2x4

= 8

f(3) = 2x4^(3-1)

= 2x4^2

= 2.16

= 32

= 2x6 ^(1-1)

= 2x1

= 2

f(2) = 2x6^(2-1)

= 2x6

= 12

f(3) = 2x6^(3-1)

= 2x6^2

= 2x36

= 72

Snezhnost [94]3 years ago

5 0

Answer:4

Step-by-step explanation:

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2 years ago

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

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Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

8 0

3 years ago

Read 2 more answers

Coefficiants of (2x+y)^4

sattari [20]

By the binomial theorem,

$(2x+y)^4=\displaystyle\sum_{k=0}^4\binom 4k(2x)^{4-k}y^k=\sum_{k=0}^4\binom 4k2^{4-k}x^{4-k}y^k$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

Then the coefficients of the $x^{4-k}y^k$ terms in the expansion are, in order from $k=0$ to $k=4$ ,

$\dbinom 402^{4-0}=1\cdot2^4=16$

$\dbinom412^{4-1}=4\cdot2^3=32$

$\dbinom422^{4-2}=6\cdot2^2=24$

$\dbinom432^{4-3}=4\cdot2^1=8$

$\dbinom442^{4-4}=1\cdot2^0=1$

3 0

3 years ago