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3 years ago

I need help with how to find 2 equations for thy problem

Mathematics

1 answer:

Mrrafil [7]3 years ago

4 0

To know how to find the answer you these two equations you figure out what the line is solid or dash.
Then you find the slope y-int form for the first equation and repeat for the second.
For your equation it would be y > 2x+2 and the second equation would be y> or equal to -1x+0

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Set up the equations to determine x and y.. help please

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Answer:

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3 years ago

How do you get the game among us on hp laptop

earnstyle [38]

Answer:

You have to pay for it

Step-by-step explanation:

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2 years ago

Read 2 more answers

Can anyone help me!

zhenek [66]

Answer:

$\large \text{$ \sf 2^{-64}$}$

Step-by-step explanation:

Given:

$\large \text{$ \sf (256)^{-4^{\frac{3}{2}}}$}$

This reads as: 256 to the power of "-4 to the power of 3/2".

Therefore, we need to deal with the "-4 to the power of 3/2" first.

Rewrite the 4 as 2²:

$\large \text{$ \sf \implies -(2^2)^{\frac{3}{2}}$}$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\large \text{$ \sf \implies -2^{\left(2 \times \frac{3}{2}\right)}$}$

$\large \text{$ \sf \implies -2^{3}$}$

Therefore:

$\large \text{$ \sf \implies -2^3=(-2)(-2)(-2) = -8$}$

Replace "-4 to the power of 3/2" with -8 :

$\large \text{$ \sf \implies 256^{-8}$}$

Rewrite 256 as 2⁸ :

$\large \text{$ \sf \implies (2^8)^{-8}$}$

$\textsf{Again, apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\large \text{$ \sf \implies 2^{(8 \times -8)}$}$

$\large \text{$ \sf \implies 2^{-64}$}$

In one complete calculation:

$\large\begin{aligned} \sf (256)^{-4^{\frac{3}{2}}} & = \sf (256)^{-(2^2)^{\frac{3}{2}}}\\& = \sf (256)^{-2^{\left(2 \times \frac{3}{2}\right)}}\\& =\sf (256)^{-2^{3}}\\& = \sf (256)^{-8}\\& \sf = (2^8)^{-8}\\& = \sf 2^{(8 \times -8)}\\& =\sf 2^{-64}\end{aligned}$

Learn more about exponent rules here:

brainly.com/question/28211147

brainly.com/question/27959936

4 0

1 year ago

Read 2 more answers

Jason went to the post office and bought both 41-cent stamps and 26-cent postcards and spent $10.28. The number of stamps was 4

Charra [1.4K]

<h3>Jason bought 20 stamps of $0.41 each and 8 postcards of $0.26 each.</h3>

<em><u>Solution:</u></em>

Let stamps be s and postcards be p

Given that,

The number of stamps was 4 more than twice the number of postcards

s = 4 + 2p -------- eqn 1

Jason bought both 41-cent stamps and 26-cent postcards and spent $10.28

41 cent = $ 0.41

26 cent = $ 0.26

Therefore,

$s \times 0.41 + p \times 0.26 = 10.28$

0.41s + 0.26p = 10.28 --------- eqn 2

Substitute eqn 1 in eqn 2

0.41(4 + 2p) + 0.26p = 10.28

1.64 + 0.82p + 0.26p = 10.28

1.08p = 10.28 - 1.64

1.08p = 8.64

Divide both sides by 1.08

p = 8

Substitute p = 8 in eqn 1

s = 4 + 2(8)

s = 4 + 16

s = 20

Thus Jason bought 20 stamps and 8 post cards

8 0

3 years ago

Terry Barlett worked from 7:00 am to 10:45 am and from 12:30 pm to 3:15 pm at a chocolate shop. Find the total hours

irakobra [83]

Answer:

6 hours 30 minutes

Step-by-step explanation:

morning work=3 hours 45 minutes

afternoon work=2 hours 45 minutes

total hours=?

Here,

3 hours 45 minutes+2 hours 45 minutes

=6 hours 30 minutes

Therefore,Terry Baelett worked 6hours 30 minutes a day.

6 0

2 years ago