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hammer [34]
4 years ago
8

1. This is the graph of which equation? You may want to use x and y intercepts to help you decide.

Mathematics
1 answer:
stepan [7]4 years ago
8 0
1. <span>5x + 5y = 25
2. </span><span>4x + y = 5
3. </span><span>4x - 6y = 24</span>
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Could you help me with this page
NeTakaya

5. area 0.75*2*0.75*2=2.25 m²

scale 1 in.=2 m so

8 0
3 years ago
Create a pattern with the rule. -4.
Bumek [7]
One pattern is 22, 18, 14, 10, 6, 2 hope this helps just keep subtracting 4. You can go from the highest number to the lowest. Thanks for asking!
7 0
3 years ago
In a drawer there are 6 black socks,8 white socks,2 red socks, and 4 blue socks. A boy takes one socks from the drawer, without
galina1969 [7]

Answer:

1/5

Step-by-step explanation:

Chance is also referred to as probability

probability = Number of event/Total samples space

If there are 6 black socks,8 white socks,2 red socks, and 4 blue socks in a drawer, the total number of socks will be the sample space

Sample space = 6 + 8 + 2 + 4

Sample space = 20socks

Total number of blue socks is the event

Number of event = 4 blue socks

Chance that the socks is blue = n(E)/n(S)

Chance that the socks is blue = 4/20

Chance that the socks is blue = 1/5

3 0
3 years ago
To find 23 times 8, double and halve
Papessa [141]
The answer is 184. 23 * 8 = 184. 184 * 2 = 368. 368 ÷ 2 = 184
6 0
3 years ago
Read 2 more answers
Express the quotient of z1 and z2 in standard form given that <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20-3%5Bcos%28%5C
Lesechka [4]

Answer:

Solution : -\frac{3}{4}-\frac{3}{4}i

Step-by-step explanation:

-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right]

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}

=-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right) ÷ 2\sqrt{2}\left(0-1\right)i

= 3\left(-\frac{\sqrt{2}i}{2}+\frac{\sqrt{2}}{2}\right) ÷ -2\sqrt{2}i

= \frac{3\left(1-i\right)}{\sqrt{2}}÷ 2\sqrt{2}i = -3-3i ÷ 4 = -\frac{3}{4}-\frac{3}{4}i

As you can see your solution is the last option.

3 0
3 years ago
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