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3 years ago

The pier is 200 meters long. each board is 80 centimeters wide. how many boards are in the pier?

Mathematics

1 answer:

Artemon [7]3 years ago

6 0

100 cm = 1 meter
80 cm = 0.8 meter
200 meters/.8 meters = number of boards
250

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2 years ago

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3 years ago

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Mrs.scott has 30 pairs of shoes.Out of the 30 pairs 24 of them are black.what percent of shoes are black

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3 years ago

Which equation has infinite solutions?

AnnyKZ [126]

The equation that has an infinite number of solutions is $2x + 3 = \frac{1}{2}(4x + 2) + 2$

<h3>How to determine the equation?</h3>

An equation that has an infinite number of solutions would be in the form

a = a

This means that both sides of the equation would be the same

Start by simplifying the options

3(x – 1) = x + 2(x + 1) + 1

3x - 3 = x + 3x + 2 + 1

3x - 3 = 4x + 3

Evaluate

x = 6 ----- one solution

x – 4(x + 1) = –3(x + 1) + 1

x - 4x - 4 = -3x - 3 + 1

-3x - 4 = -3x - 2

-4 = -2 ---- no solution

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

2x + 3 = 2x + 1 + 2

2x + 3 = 2x + 3

Subtract 2x

3 = 3 ---- infinite solution

Hence, the equation that has an infinite number of solutions is $2x + 3 = \frac{1}{2}(4x + 2) + 2$

Read more about equations at:

brainly.com/question/15349799

#SPJ1

<u>Complete question</u>

Which equation has infinite solutions?

3(x – 1) = x + 2(x + 1) + 1

x – 4(x + 1) = –3(x + 1) + 1

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

$\frac 13(6x - 3) = 3(x + 1) - x - 2$

5 0

2 years ago

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

2 years ago