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3 years ago

The perimeter of a retangular pool is 360 inches the length is 8 inches longer than the width find the dimensions of the pool

1 answer:

4 0

360=2w+2l
l=8+w

Substitute for l
360=2w+2(8+w)
360=2w+16+2w
360=4w+16
344=4w
86=w

Plug in the w value
l=8+(86)
l=94

Final answer: Length-94 inches, Width-86 inches

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suppose that a biologist is watching a trail known for wildebeest migration. During the first minute, 24 wildebeests migrated pa

Monica [59]

Answer:

=1050 wildebeests

Step-by-step explanation:

We can form an arithmetic series for the wildebeest migration.

Sₙ=n/2(2a+(n-1)d) where n is the number of terms, d is the common difference and a is the first term.

a=24

n=20

d=3

Sₙ=(20/2)(2(24)+(20-1)3)

Sₙ=10(48+57)

=1050 wildebeests

6 0

4 years ago

The distance from Earth to the moon is 384,400 kilometers. What is this distance expressed in scientific notation?

Oksi-84 [34.3K]

Answer:

3.844 x 10 to the power of 5 is the answer i think

Step-by-step explanation:

4 0

3 years ago

I need helppppppp plzzzzzz

Zanzabum

14 x 10^3 = 14,000 (the elephants weight)

So the cat weighs 14 pounds

3 0

4 years ago

Read 2 more answers

Please walk me through how to do this so I can di the other questions

Naddik [55]

Answer:

Axis is a vertical line at x = 2

Vertex is (2, -1)

y-intercept is (0, 3)

Solutions are x = 1 and x = 3

Step-by-step explanation:

To draw the graph of the quadratic equation you must find at least 5 points lie on the graph by choose values of x and find their values of y

Let us do that

Use x = -1, 0, 1, 2, 3, 4, 5

∵ y = x² - 4x + 3

∵ x = -1

∴ y = (-1)² - 4(-1) + 3 = 1 + 4 + 3 = 8

→ Plot point (-1, 8)

∵ x = 0

∴ y = (0)² - 4(0) + 3 = 0 + 0 + 3 = 3

→ Plot point (0, 3)

∵ x = 1

∴ y = (1)² - 4(1) + 3 = 1 - 4 + 3 = 0

→ Plot point (1, 0)

∵ x = 2

∴ y = (2)² - 4(2) + 3 = 4 - 8 + 3 = -1

→ Plot point (2, -1)

∵ x = 3

∴ y = (3)² - 4(3) + 3 = 9 - 12 + 3 = 0

→ Plot point (3, 0)

∵ x = 4

∴ y = (4)² - 4(4) + 3 = 16 - 16 + 3 = 3

→ Plot point (4, 3)

∵ x = 5

∴ y = (5)² - 4(5) + 3 = 25 - 20 + 3 = 8

→ Plot point (5, 8)

→ Join all the points to form the parabola

From the graph

∵ The axis of symmetry is the vertical line passes through the vertex point

∵ x-coordinate of the vertex point is 2

∴ Axis is a vertical line at x = 2

∵ The coordinates of the vertex point of the parabola are (2, -1)

∴ Vertex is (2, -1)

∵ The parabola intersects the y-axis at point (0, 3)

∴ y-intercept is (0, 3)

∵ x² - 4x + 3 = 0

∵ The solutions of the equation are the values of x at y = 0

→ That means the intersection points of the parabola and the x-axis

∵ The parabola intersects the x-axis at points (1, 0) and (3, 0)

∴ Solutions are x = 1 and x = 3

3 0

3 years ago

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

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5 0

3 years ago

Read 2 more answers