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3 years ago

Oscar ran the 100-yard dash in 12.58 seconds. Jesiah ran the 100-yard dash in 11.59 seconds.

Mathematics

2 answers:

levacccp [35]3 years ago

8 0

Answer:

0.99

Step-by-step explanation:

marysya [2.9K]3 years ago

7 0

0.99 because 12.58-11.59=0.99

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PLEASE HELP!!

erik [133]

$2.=c.\\\\2\sin4x\cos4x=2\sin(2\cdot4x)=2\sin8x\\\\Used:\\\sin2\alpha=2\sin\alpha\cos\alpha$

$1.=b.\\\\\csc x-\sin x=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{\cos^2x}{\sin x}\\\\=\dfrac{\cos x\cos x}{\sin x}=\cos x\cdot\dfrac{\cos x}{\sin x}=\cos x\cot x\\\\Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\cot x=\dfrac{\cos x}{\sin x}$

$3.=a.\\\\\dfrac{\sin x-1}{\sin x+1}=\dfrac{\sin x-1}{\sin x+1}\cdot\dfrac{\sin x+1}{\sin x+1}=\dfrac{\sin^2x-1^2}{(\sin x+1)^2}=\dfrac{\sin^2x-1}{(\sin x+1)^2}\\\\=\dfrac{-(1-\sin^2x)}{(\sin x+1)^2}=\dfrac{-\cos^2x}{(\sin x+1)^2}\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x$

8 0

3 years ago

Find the solution to the initial value problem

Karo-lina-s [1.5K]

Answer:

y = (11x + 13)e^(-4x-4)

Step-by-step explanation:

Given y'' + 8y' + 16 = 0

The auxiliary equation to the differential equation is:

m² + 8m + 16 = 0

Factorizing this, we have

(m + 4)² = 0

m = -4 twice

The complimentary solution is

y_c = (C1 + C2x)e^(-4x)

Using the initial conditions

y(-1) = 2

2 = (C1 -C2) e^4

C1 - C2 = 2e^(-4).................................(1)

y'(-1) = 3

y'_c = -4(C1 + C2x)e^(-4x) + C2e^(-4x)

3 = -4(C1 - C2)e^4 + C2e^4

-4C1 + 5C2 = 3e^(-4)..............................(2)

Solving (1) and (2) simultaneously, we have

From (1)

C1 = 2e^(-4) + C2

Using this in (2)

-4[2e^(-4) + C2] + 5C2 = 3e^(-4)

C2 = 11e^(-4)

C1 = 2e^(-4) + 11e^(-4)

= 13e^(-4)

The general solution is now

y = [13e^(-4) + 11xe^(-4)]e^(-4x)

= (11x + 13)e^(-4x-4)

3 0

3 years ago

Which of the following equations could be solved for a radius? Select all that apply.

zalisa [80]

Answer:

A = π · (r²)

Step-by-step explanation:

$A =$ π · r² is the area of a circle.

While π · r² · h can also give you the radius, it can only do so for the Volume $V$ , not the Area $A$ .

$V=lwh$ doesn't really apply for a circular object, as it requires the length and width. For circular objects, both are equal to the diameter of the object, and 2² · r² · h does not equal the Volume.

$V=$ π · r³ seems awfully like the volume of a sphere, but there's something missing. The true volume of a sphere is $V=\frac{4}{3}$ · π · r³, not $V=$ π · r³.