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3 years ago

What is the value of f(-3) when f(x)=5(2)x

Mathematics

1 answer:

xeze [42]3 years ago

4 0

Answer:

-30

Step-by-step explanation:

f(x)=5(2)x

f(-3)=5(2)(-3)

f(-3)=10(-3)

f(-3)=-30

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Write the expression that shows " 3 times the sixth power of 10."

Fed [463]

You would write a 3 with a little 6 to the top right of the number or on the computer as 3^6

8 0

3 years ago

the capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of metal sheet would be needed

lord [1]

Answer:

7.7 square metres or 7.669 square metres of metal sheet

Step-by-step explanation:

Volume of cylinder = 2×22/7×r×100cm

15.4 litres = 15,400 cm cube

Radius = 15 400×7/22 × 50

= 98cm

Surface area of cylinder = (2×22/7×49^2) + (2×22/7×49 × 100)

= 15 092 + 61 600

= 76 692 square cm

= 7.669 square metres of metal sheet required

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.