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2 years ago

3. A children's slide is 10 feet long and inclines 64° from the ground. The

Mathematics

1 answer:

Nata [24]2 years ago

3 0

Answer:

87.0°

Step-by-step explanation:

The law of sines can be used to solve this. We have two sides of a triangle and the angle opposite one of them. We want to find the angle opposite the other known side.

In the attached, the triangle is ΔACS. We have side "a" = 9, and side "c" = 10. Angle A is given as 64°. The law of sines tells us ...

sin(C)/c = sin(A)/a

sin(C) = (c/a)sin(A)

C = arcsin((c/a)sin(A)) = arcsin(10/9·sin(64°)) ≈ 87.03°

The ladder makes an angle of about 87° with the ground.

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Completely factor the polynomial. 16x 2 - 25

Sunny_sXe [5.5K]

Answer:

(4x-5)(4x+5)

Step-by-step explanation:

16x^2 - 25

Rewriting

(4x)^2 - 5^2

We recognize that this is the difference of squares

a^2 - b^2 = (a-b) (a+b)

(4x)^2 - 5^2 = (4x-5)(4x+5)

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2 years ago

A football team loses 4 yards on each 3 consecutive plays. Find the total change in yard from where the team started

BigorU [14]

Answer:

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Step-by-step explanation:

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2 years ago

The perimeter of a rectangular painting is 270 cm.

Ainat [17]

The anwser is 53 cm each

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2 years ago

Read 2 more answers

Use suitable identities to find the following products <br><br>(x-5) (x-5)

Marrrta [24]

Step-by-step explanation:

$(x - 5)(x - 5) \\ {x}^{2} - 5x - 5x + 25 \\ {x}^{2} - 10x + 25$

I hope it makes sense

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2 years ago

hus, D satisfying (ABC)DequalsI exists. Why does the expression for D found above also satisfy D(ABC)equalsI, thereby showing

Assoli18 [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

First Question

Option A is correct

Second Question

Option C is correct

Third Question

$D = A^{-1} * B^{-1} * C^{-1}$

Fourth Question

So substituting for D in (ABC) D = I

$(ABC) * A^{-1} * B^{-1} * C^{-1} = I$

$I = I$

This proof that ABC is invertible

Step-by-step explanation:

From the question we are told that

A , B and C are invertible which means that $A^{-1} , B^{-1}, C^{-1}$ exist

Now

From the question

(ABC) D = I

Where I is an identity matrix

Now when we multiply both sides by $A^{-1}$ we have

$A^{-1} A BCD = A^{-1} * I$

$IBCD = A^{-1}$

Now when we multiply both sides by $B^{-1}$ we have

$B^{-1 } *I BCD = A^{-1} * B^{-1}$

$I CD = A^{-1} * B^{-1}$

Now when we multiply both sides by $C^{-1}$ we have

$C^{-1} * I CD = A^{-1} * B^{-1} * C^{-1}$

$I D = A^{-1} * B^{-1} * C^{-1}$

$D = A^{-1} * B^{-1} * C^{-1}$

So substituting for D in the above equation

$(ABC) * A^{-1} * B^{-1} * C^{-1} = I$

$I = I$

This proof that ABC is invertible

4 0

2 years ago