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3 years ago

I don’t know what words to use

Mathematics

1 answer:

irinina [24]3 years ago

5 0

Hey there!

I'll answer these questions, but first, I can explain what the formula is:

y = a|x-h| + k

A regulates how wide or narrow the graph is

H tells you how much left or right

K tells you how much Up and down

(h,k) is the vertex.

1) The graph is of standard shape. A is 1.

2) It opens up.

3) Y values increase while x values increase, or y values increase as x values decrease (infinitely).

4) x- (-1,0) and (-3,0) y- (1,0)

5) Special point could be the vertex, which is (-2, -1)

Hope this helps!

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You have to prove it!<br>EVEN IF YOU KNOW ONE OF THEM PLZZZ HELPPP!:(<br>.

natita [175]

Answer:

xi.) 1 = 1

xii.) 1 + 1 = 2

Step-by-step explanation:

see explanation in attached picture,

i hope it helped.

7 0

3 years ago

The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am

posledela

Step-by-step walkthrough:

Well a standard half-life equation looks like this.

$N = N_0 * (\frac{1}{2})^{t/p$

$N_0$ is the starting amount of parent element.

$N$ is the end amount of parent element

$t$ is the time elapsed

$p$ is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

$N(t) = 74g * (\frac{1}{2})^{t/2.8days$

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

$N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2} = 37g$

Now we just gotta do some algebra. Use the original function but this time, replace $N(t)$ with 10g and solve algebraically.

$10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}$

Take the log of both sides.

$log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})$

Use the exponent rule for log laws that, $log(b^x) = x*log(b)$

$log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})$

$\frac{log(\frac{5}{37})}{log(\frac{1}{2})} = \frac{t}{2.8days}$

$2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})} = t$

slap that in your calculator and you get

$t = 8.1 days$

7 0

2 years ago

The diameter, D, of a sphere is 18.8m. Calculate the sphere volume, V. Use the value 3.14 for pi

mamaluj [8]

Answer:

3.14159369398579323..... and so on

7 0

3 years ago

The slope of the line?

Ostrovityanka [42]

The slope of this line is 3/2.

5 0

4 years ago

The expression was rewritten using the ( blank?) <br> i’m so confused pls help

RSB [31]

I am solving the equation in simplest way

Line:-1

$\\ \tt\bull\longmapsto 9.(7+1)$

Solve the bracket

$\\ \tt\bull\longmapsto 9(8)$

$\\ \tt\bull\longmapsto 72$

Line:-2

$\\ \tt\bull\longmapsto 9.7+9.1$

$\\ \tt\bull\longmapsto 63+9$

$\\ \tt\bull\longmapsto 72$

3 0

3 years ago

Read 2 more answers