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3 years ago

7

Last week, labron jogged for 2 1/4 miles. This week , he plants to jog 2 times as far as last week. How many miles does Lebron P

lant to jog this week?

2 answers:

Shtirlitz [24]3 years ago

7 0

Answer:

4.5 miles

Step-by-step explanation:

2 times as far as last week

last week was 2.25 miles

2* 2.25 = 4.5 miles

prisoha [69]3 years ago

5 0

Answer: 2 times 2 1/4 is 9/4 times 2 wich is 18/4 simplified is 4 1/2

Step-by-step explanation:

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Write an equation in point-slope form, if m= 3 and passes through (4, 5).

Sergio039 [100]

Answer:

y - 5 = 3(x - 4)

General Formulas and Concepts:

<u>Algebra I</u>

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate

y₁ - y coordinate

m - slope

Step-by-step explanation:

<u>Step 1: Define</u>

Slope <em>m</em> = 3

Point (4, 5)

<u>Step 2: Write equation</u>

Point-Slope Form: y - 5 = 3(x - 4)

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What is the value of 'n' in the equation<br> -1/2(2n+4)+6=-9+4(2n+1)

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Step-by-step explanation:

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3 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:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

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

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ANSWER PLEASE QUICK!! I AM ON A TIMERRR

Fynjy0 [20]

Answer:

smaller x is to the power of 2 3x is the larger.

Step-by-step explanation:

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3 years ago

A mass oscillating up and down on the bottom of a spring can be modeled as harmonic motion. if they weight is displaced a maximu

snow_tiger [21]

9514 1404 393

Answer:

1.63 cm (across the centerline from release)

Step-by-step explanation:

If we assume time starts counting when we release the weight from its fully-extended downward position, then the position at 1.15 seconds can be found from ...

h(t) = -7cos(2πt/4)

h(1.15) = -7cos(π·1.15/2) = -7(-0.233445) ≈ 1.63412 . . . cm

That is, 1.15 seconds after the weight is released from below the resting position, it will be 1.63 cm above the resting position.

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If it is released from <em>above</em> the resting position, it will be 1.63 cm <em>below</em> the resting position at t=1.15 seconds.

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3 years ago