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3 years ago

The fraction which is not equal to 4/5 is? a)40/50 b)12/15 c)16/20 d)9/15

Mathematics

1 answer:

nikitadnepr [17]3 years ago

8 0

D because 4/5 ≠ 9/15; 5x3=15 but 4x3=12 so answer is d

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A square pyramid is 6 feet on each side. The height of the pyramid is 6 feet. What is the total area of the pyramid?

Fed [463]

Answer:

The Area of square pyramid is 36 + 36 $\sqrt{5}$ .

Step-by-step explanation:

Given as :

For A square pyramid

The measure of each base side = b = 6 feet

The height of the pyramid = h = 6 feet

Let The Total area of pyramid = A square feet

Now, According to question

<u>The area of square pyramid </u>

Area = b² + 2 × b $\sqrt{\frac{b^{2} }{4}+h^{2} }$

Where b is the square base edge

And h is the pyramid height

I.e A = 6² + 2 × 6 $\sqrt{\frac{6^{2} }{4}+6^{2} }$

Or, A = 36 + 12 × $\sqrt{\frac{36 }{4}+36 }$

Or, A = 36 + 12 × $\sqrt{9+36}$

Or, A = 36 + 12 × $\sqrt{45}$

Or, A = 36 + 12 × 3 $\sqrt{5}$

Or, A = 36 + 36 $\sqrt{5}$

So,The Area of square pyramid = A = 36 + 36 $\sqrt{5}$

Hence, The Area of square pyramid is 36 + 36 $\sqrt{5}$ . Answer

6 0

3 years ago

If a 1/4 of a whailing trip equals $12,000, what does a 1/8 shate equal

Leona [35]

Answer:

$24,000

Step-by-step explanation:

Because if it says 1/4 to 1/8 you have to double the cost

7 0

3 years ago

Read 2 more answers

I suck at geometry lol

Pavlova-9 [17]

Answer:

Step-by-step explanation:

Pythagorean theorem states that a^2 + b^2 = c^2. Let the length of the legs be x, so 2x^2 = c^2. c is the hypotenuse, which means that 2x^2 = 64. Divide both sides by 2, and we get x^2 = 32. Square root both sides and we get sqrt 32 or 4 * sqrt 2.

6 0

3 years ago

Round 4.25 to one decimal place

hammer [34]

4.3 because 5 rounds up, and 25 rounds to 30.

4 0

3 years ago

Use the following vectors to answer parts (a) and (b). v1equals=[Start 3 By 1 Matrix 1st Row 1st Column 1 2nd Row 1st Column n

erik [133]

Answer:

(1) No matter what's the value of $h$ , $\vec{v}_3$ is never in the span of $\vec{v}_1$ and $\vec{v}_2$ .

(2) The three vectors $\vec{v}_1$ , $\vec{v}_2$ , and $\vec{v}_3$ are always linearly dependent for all real $h$ .

Step-by-step explanation:

If $\vec{v}_3$ is in the span of $\vec{v}_1$ and $\vec{v}_2$ , there need to exist real $a$ and $b$ such that

$a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3$ .

Assume that such $a$ and $b$ do exist.

In other words,

$\displaystyle a \left[\begin{array}{c}{1 \\ -4\\2} \end{array}\right] + b\left[\begin{array}{c}{-4 \\ 16\\-8}\end{array}\right] = \left[\begin{array}{c}5 \\7 \\ h\end{array}\right]$ .

$\displaystyle \left[\begin{array}{c}{a \\ -4a\\2a} \end{array}\right] + \left[\begin{array}{c}{-4b \\ 16b\\-8b}\end{array}\right] = \left[\begin{array}{c}5 \\7 \\ h\end{array}\right]$ .

$\left\{\begin{array}{rrcr} a & - 4b &=& 5\\ -4a & + 16b &= &7\\2a & -8b & =&h\end{aligned}\right.$ .

Rewrite as an augmented matrix and row-reduce:

$\displaystyle \left[ \begin{array}{cc|c} 1 & -4 & 5 \\ -4 & 16 & 7 \\ 2 & -8 & h\end{array}\right]$ .

(Add four times row one to row two and $-2$ times row one to row three.)

$\displaystyle \sim \left[ \begin{array}{cc|c} 1 & -4 & 5 \\ 0 & 0 & 27 \\ 0 & 0 & h - 10\end{array}\right]$ .

Note that in row two,

Left-hand side: $0$ ;
Right-hand side: $27\neq 0$ .

In other words, this system is inconsistent. There's no real $a$ and $b$ that would satisfy the condition

$a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3$ .

Hence $\forall h \in \mathbb{R}, \quad \vec{v}_3\not \in \text{Span}\{\vec{v}_1, \vec{v}_2\}$ .

There's no real $h$ that allows $h$ , $\vec{v}_3$ to be part of the span of $\vec{v}_1$ and $\vec{v}_2$ .

If the three vectors are linearly dependent, at least one of them can be expressed as the linear combination of the other two.

Note that

$\vec{v}_2 = (-4)\vec{v}_1 + 0 \; \vec{v}_3$ . In other words, $\vec{v}_2$ can be written as the linear combination of the other two vectors. Additionally, since the coefficient in front of $\vec{v}_3$ is zero, neither the exact value of $\vec{v}_3$ nor the value of $h$ will make a difference. Therefore, for all $h \in \mathbb{R}$ , the three vectors $\vec{v}_1$ , $\vec{v}_2$ , and $\vec{v}_3$ are linearly dependent.

8 0

3 years ago