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2 years ago

The length of a swimming pool is 8 m longer than it's width and the area is 105m²

Mathematics

2 answers:

cestrela7 [59]2 years ago

7 0

Answer:

Below

Step-by-step explanation:

The area of the swimming pool is

● A = L × w where L is the length and w is the width

The length is 8 m more than the width L = w + 8

● A = (w+8)×w

● A = w^2 +8w

● 105 = w^2 + 8w

● w^2 + 8w - 105 = 0

This is a quadratic equation.

We will solve by graphing it using a graphing tool

The solutions are the intersection points with the x-axis

We will only take the positive solution since the width can't be negative.

The solution is 7 (picture below)

● w = 7 m

The length is

● L = w + 8

● L = 7 + 8

● L = 15 m

Art [367]2 years ago

6 0

width= 7m

Step-by-step explanation:

Area= L×W

L= (x+8)m

W=8m

105m²= (x+8)m × (x)m

105m²= x² + 8x

-x² - 8x + 105= 0

(x-7) (x+15) = 0

x= 7 x= -15

x=width = 7m

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Which of the following equations is the formula of f(x) = x^1/3 but shifted 4 units to the left and 4 units up?

nlexa [21]

Answer:

Hence correct chcie is C.

$f\left(x\right)=(x+4)^{\frac{1}{3}}+4$

Step-by-step explanation:

Given function is $f\left(x\right)=x^{\frac{1}{3}}$

Now it says that function is shifted 4 units to the left and 4 units up.

We need to find about which of the given choice is correct for the given transformation.

When f(x) is shifted "h" units left then we write f(x+h)

So $f\left(x\right)=x^{\frac{1}{3}}$ will change to

$f\left(x\right)=(x+4)^{\frac{1}{3}}$

When f(x) is shifted "h" units up then we write f(x)+h

So $f\left(x\right)=(x+4)^{\frac{1}{3}}$ will change to

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4 0

2 years ago

Read 2 more answers

Simplify each sum or difference. State any restrictions on the variables.

Dvinal [7]

Answer:

2(d-2)/(2d+1) or 1 - 5/(2d+1)

d can not be -½

Step-by-step explanation:

(d-3)/(2d+1) + (d-1)/(2d+1)

Lcm: 2d+1

(d-3+d-1)/(2d+1)

= (2d-4)/(2d+1)

= 2(d-2)/(2d+1)

If further simplification required:

= (2d-4)/(2d+1)

= (2d+1-5)/(2d+1)

= (2d+1)/(2d+1) - 5/(2d+1)

= 1 - 5/(2d+1)

2d+1 = 0

When d = -½

So d can not be -½ because when the denominator becomes 0, fraction becomes undefined

7 0

2 years ago

What is the solution to this equation?

Usimov [2.4K]

-0.5 n^2=-18
n^2= 36
n= sqrt (36)=6 or -6

Hope this can help.

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2 years ago

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Could someone please help me? Thank you and explain the work because I don’t get this

lesya [120]

Answer:

5.59 times per second.

Step-by-step explanation:

Direct variation is in the form:

$y=kx$

Where k is the constant of variation.

Inverse variation is in the form:

$\displaystyle y=\frac{k}{x}$

In the given problem, the frequency of a vibrating guitar string varies inversely as its length. In other words, using f for frequency and l for length:

$\displaystyle f=\frac{k}{\ell}$

We can solve for the constant of variation. We know that the frequency f is 4.3 when the length is 0.65 meters long. Thus:

$\displaystyle 4.3=\frac{k}{0.65}$

Solve for k:

$k=4.3(0.65)=2.795$

So, our equation becomes:

$\displaystyle f=\frac{2.795}{\ell}$

Then when the length is 0.5 meters, the frequency will be:

$\displaystyle f=\frac{2.795}{.5}=5.59\text{ times per second.}$

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2 years ago

PPPLLLZZZZZZ HELP! TnT ASAP! ( MATH ) FIRST ONE GETS BRAINLIST, A THANK YOU AND POINTS!!

spin [16.1K]

Answer:

1).

Alyssa:

About 1 book

Brandon:

About 1 book

Alyssa:

7/12 of a book

Brandon:

4/5 of a book

Step-by-step explanation:

Alyssa:

1/4=1/2

1/3=1/2

1/2+1/2=1

1/4= 3/12

1/3=4/12

3/12+4/12=7/12

Brandon:

2/5=1/2

4/10=1/2

1/2+1/2=1

2/5=4/10

4/10+4/10=8/10=4/5

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2 years ago

The length of a swimming pool is 8 m longer than it's width and the area is 105m²​

The length of a swimming pool is 8 m longer than it's width and the area is 105m²