<img src="https://tex.z-dn.net/?f=b%20%3D%20%20log_%7B10%7D%28a%29%20" id="TexFormula1" title="b = log_{10}(a) " alt="b = log

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jeyben [28]

3 years ago

{10}(a) " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

exis [7]3 years ago

5 0

Answer:

$b = log_{10}(a) \\ \\ \therefore \: {10}^{b} = a$

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Three muffin pans with six in each pan what is the fraction for this

irakobra [83]

Answer:

1/18 ?

Step-by-step explanation:

I'm not sure if you're written the whole question?

3 pans with 6 muffins in each pan, so total number of muffins = 3 x 6 = 18

So each muffin is 1/18

4 0

2 years ago

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Which is the area of the rectangle?<br> 69<br> 115<br> 100

Burka [1]

Answer:

Step-by-step explanation:

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3 years ago

Which phrase best describes the translation from the graph y=(x-5)^2+7 to the graph of y=(x+1)^2-2

Helga [31]

Start from the parent function $f(x)=x^2$

In the first case, you are computing

$f(x-5)+7$

In the second case, you are computing

$f(x+1)-2 /tex] There are two translation going on: when you transform [tex] f(x) \to f(x+k)$ , you translate the function horizontally, $k$ units left if $k>0$ and $k$ units right if $k$ .

On the other hand, when you transform $f(x) \to f(x)+k$ , you translate the function vertically, $k$ units up if $k>0$ and $k$ units down if $k$ .

So, the first function is the "original" parabola $f(x)=x^2$ , translated $5$ units right and $7$ units up. Likewise, the second function is the "original" parabola $f(x)=x^2$ , translated $1$ units left and $2$ units down.

So, the transformation from $(x-5)^2+7$ to $(x+1)^2-2$ is: go $6$ units to the left and $2$ units down

8 0

3 years ago

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Which of the following is equal to the rational expression when x -2 or -1? x^2-4/(x+2)(x+1)

Orlov [11]

Answer:

(x+2)(x+1)

Step-by-step explanation:

Assuming im interpreting this correctly, it should be (x+2)(x+1), since if you plug in -2 and -1, it will give you 0, which is the answer.

8 0

3 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago