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4 years ago

Need help!!!!!!!!!

Mathematics

1 answer:

victus00 [196]4 years ago

4 0

It is not a function because the -2 x-values have two different y-values.

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Pleasee someoneee help asapp

xxMikexx [17]

A is a possibility but I'm not 100% sure

5 0

3 years ago

9a - 2 = -2 help please

Svetach [21]

$\text {Hello! Let's Solve this Problem!}$

$\text {\underline {The First Step is to Add 2 to Both Sides}}$

$\text {-2+2=0 (this crosses out)}\\\text {-2+2=0}$

$\text {\underline {The Final Step is to Divide Both Sides by 9}}$

$\text {9a/9=0 (this crosses out. Leaving a by itself)}\\\text {0/9=}$

$\text {Your Answer Would Be:}$

$\huge\boxed {a=0}$

$\text {Best of Luck!}$

3 0

3 years ago

Read 2 more answers

How many half-lives of radon-222 have passed in 11.46 days? If 5.2 × 10−8 g of radon-222 remain in a sealed box after 11.46 days

Svetach [21]

Answer:

(i) Approximately 3 half lifes

(ii) $4.21\times 10^{-7}\text{ g}$

Step-by-step explanation:

(i) ∵ The half life of Radon-222 is approximately 3.8 days,

So, the number of half life in 11.46 days = $\frac{11.46}{3.8}$ ≈ 3

(ii) Since, the half life formula is,

$N=N_0 (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

Where,

$N_0$ = initial quantity,

t = number of periods

$t_{\frac{1}{2}}$ = half life of the quantity,

Given,

N = $5.2\times 10^{-8}\text{ g}$

t = 11.46 days,

$t_{\frac{1}{2}} = 3.8\text{ days}$

$\implies 5.2\times 10^{-8}=N_0 (\frac{1}{2})^\frac{11.46}{3.8}$

$\implies N_0=2^{\frac{11.46}{3.8}}\times 5.2\times 10^{-8}\approx 4.21\times 10^{-7}\text{ g}$

7 0

4 years ago

Can anyone help me with this?

yawa3891 [41]

Answer:

Below.

Step-by-step explanation:

7) c^2 = a^2 + b^2

15^2 = 11^2 + b^2

b^2 = 225 - 121 = 104

b = √104

= √4√26

= 2√26.m

8). 4^2 = a^2 + (√6)^2

a^2 = 16 - 6

a^2 = 10

a = √10.

7 0

2 years ago

Which one is bigger?

Alex777 [14]

Answer:

Step-by-step explanation:

$\frac{4}{5} ^{2}$ = 0.64

$\frac{3}{5} / \frac{4}{5}$ = 0.75

Therefore, the answer is B.

Hope this helps!

3 0

2 years ago