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2 years ago

Does this graph represent a function

Mathematics

1 answer:

Ilia_Sergeevich [38]2 years ago

7 0

Hi!

A function is when you can look at any number on the graph, and it only has one value in the line. For example, look at 2 on the y-axis. It only has one point on the line, but -2 on the y-axis has two points on the line. I hope that makes sense.

Therefore the answer is B. No, because some of the y-values are paired with 2 x-values.

Hope this helps! :)

-Peredhel

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2 years ago

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3 years ago

.. Which of the following are the coordinates of the vertices of the following square with sides of length a?

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Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Step-by-step explanation:

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

To find the sides of a square, let us use the distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

Thus, the square with vertices $O(0,0), S(a,0), T(a,a), W(0,a)$ has sides of length a.

Thus, the correct answers are option a and option d.

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2 years ago

A cube and a square pyramid were joined to form the composite solid.

Yakvenalex [24]

We know that
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surface area of the cube=5*[12*12]----> 720 in²

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2 years ago

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Answer:

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Step-by-step explanation:

Given the 2 equations

y = x + 2 → (1)

y = x² + 7x + 10 → (2)

Since both equations express y in terms of x we can equate them, that is

x² + 7x + 10 = x + 2 ( subtract x + 2 from both sides )

x² + 6x + 8 = 0 ← in standard form

(x + 2)(x + 4) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 2 = 0 ⇒ x = - 2

x + 4 = 0 ⇒ x = - 4

Substitute these values into (1) for corresponding values of y

x = - 2 : y = - 2 + 2 = 0 ⇒ (- 2, 0 )

x = - 4 : y = - 4 + 2 = - 2 ⇒ (- 4, - 2 )

Solutions are (- 2, 0 ) and (- 4, - 2 )

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3 years ago