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3 years ago

if u have 20 square pieces of wood describe ways you could make a rectangle by placing them side by side

Mathematics

1 answer:

Llana [10]3 years ago

6 0

5 times 4 equals 20
2 times 10 equals 20

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Pls help fast !!!!!!!!!!!!!!!!!!!!!!!!

Sergeeva-Olga [200]

Answer:

$\frac{2}{3}$

Step-by-step explanation:

⅓×2

When multiplied by 2 will be;

$\frac{1}{3} \times \frac{2}{1}$

So you 2×1=2

3×1 is equal to 3

So the final answer will be;

$\frac{2}{3}$

4 0

3 years ago

Read 2 more answers

Given that A, O & B lie on a straight line segment, evaluate COD.

GrogVix [38]

(3x + 94) I’m not sure

8 0

3 years ago

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Let the graph of g be a vertical stretch by a factor of 2, followed by a translation 5 units down of the graph of f(x) = e". Wri

Burka [1]

Answer:

g(x) = 2e - 5

Step-by-step explanation:

f(x) = e

2 f(x) = 2e : vertical stretch by a factor 2

2 f(x) - 5 : transition 5 units down

g(x) = 2 f(x) - 5 = 2e - 5

6 0

3 years ago

Find the quotient. <br><br> 2x-3/x divided by 7/x^2

BlackZzzverrR [31]

$\dfrac{2x-3}{x}:\dfrac{7}{x^2}=\dfrac{2x-3}{x}\cdot\dfrac{x^2}{7}=\dfrac{2x-3}{1}\cdot\dfrac{x}{7}\\\\=\dfrac{(2x-3)\cdot x}{1\cdot7}=\dfrac{2x^2-3x}{7}$

4 0

3 years ago

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Find the value(s) of k such that 4x² + kx + 4 = 0 has 1 rational solution

Blababa [14]

Answer:

If you mean only one rational solution, the answer is

$k_1 = 8, k_2 = -8$

If you mean at least 1 rational solution, the answer is

$k\in (-\infty, -8]\cup[8, \infty)$

Step-by-step explanation:

$4x^2 + kx + 4 = 0$

Let's calculate the discriminant.

$\Delta = b^2 - 4ac$

$\Delta = k^2 -4 \cdot 4 \cdot 4$

$\Delta = k^2 -64$

Now, remember that:

$\text{If } \Delta > 0 : \text{2 Real solutions}$

$\text{If } \Delta = 0 : \text{1 Real solution}$

$\text{If } \Delta < 0 : \text{No Real solution}$

Therefore, I will just consider the first two cases.

$k^2 - 64 > 0$

and

$k^2 -64 = 0$

$\boxed{\text{For } k^2 - 64 > 0}$

$k^2 > 64 \Longleftrightarrow k>\pm\sqrt{64} \Longleftrightarrow k\in (-\infty, -8)\cup(8, \infty)$

$\boxed{\text{For } k^2 - 64 = 0}$

$k^2 = 64 \Longleftrightarrow k=\pm\sqrt{64} \implies k_1 = 8, k_2 = -8$

7 0

3 years ago