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2 years ago

7

What’s the answer

1 answer:

Dmitriy789 [7]2 years ago

4 0

The y intercept is 3

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a rectangular lawn has an area of a^3 - 125. use the difference of cubes to find out the dimensions of the rectangle.

ANEK [815]

The area of a rectangle is the product of its dimensions

The dimensions of the rectangle are: $\mathbf{Length = a -5}$ and $\mathbf{Width = a^2 + 5a + 25}$

The area is given as:

$\mathbf{Area = a^3 - 125}$

Express 125 as 5^3

$\mathbf{Area = a^3 - 5^3}$

Apply difference of cubes

$\mathbf{Area = (a - 5)(a^2 + 5a + 5^2)}$

$\mathbf{Area = (a - 5)(a^2 + 5a + 25)}$

The area of a rectangle is:

$\mathbf{Area = Length \times Width}$

So, by comparison:

$\mathbf{Length = a -5}$

$\mathbf{Width = a^2 + 5a + 25}$

Read more about areas at:

brainly.com/question/3518080

6 0

2 years ago

A plane rises from take-off and flies at an angle of 4° with the horizontal runway. When it has gained 450 feet, find the dis

jeyben [28]

Answer: a)6447feet

b) 6446.98feet

Step-by-step explanation:

For the first question, it is represented with figure 1 in the picture attached

For the second question, it is represented with figure 2 in the picture attached

3 0

3 years ago

Please help me I don’t understand

Ann [662]

Answer:

I think its C if I’m wrong sorry

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

50 points!!!!!!!!!!!!

igomit [66]

Answer: the first one .00000000058

Step-by-step explanation:

8 0

1 year ago

Read 2 more answers

Answer the question in the picture

nekit [7.7K]

Recall the angle sum identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

Now,

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$

Divide through numerator and denominator by $\cos x\cos y$ to get

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

Next, we use the fact that $x,y$ lie in the first quadrant to determine that

$\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2$

$\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}$

So we then have

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1$

Finally,

$\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73$

4 0

2 years ago