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andrey2020 [161]
3 years ago
15

Choose the multiples of 5. 10 .32.35.40 and 81​

Mathematics
2 answers:
ololo11 [35]3 years ago
3 0
10, 35 & 40 are all multiples of five
alexandr1967 [171]3 years ago
3 0
10, 35 and 40 can all be divided by five which makes them multiples of five
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What is the solution of sqrt x+2-15=-3
antiseptic1488 [7]

Answer:

Step-by-step explanation:

Without brackets, we are not exactly sure what is under the root sign. There are 3 choices.

sqrt(x) + 2 - 15 = - 3

sqrt(x + 2) - 15 = - 3

sqrt(x + 2 - 15) = - 3

I think the middle one is what you intend. If not leave a note.

sqrt(x + 2) - 15 = - 3                Add 15 to both sides.

sqrt(x + 2) - 15+15 = - 3+15     Combine

sqrt(x + 2) = 12                        Square both sides

x + 2 = 12^2                             Do the right

x + 2 = 144                               Subtract 2 from both sides.

x + 2-2 = 144-2

x = 142

3 0
4 years ago
Please help me with the answers
Illusion [34]

Answer:

slope= 3/1

y-intercept (0,2)

y= 3/1 b + 2

4 0
3 years ago
Steps to solve x/8-5=-1
topjm [15]

Answer: x = 32

Step-by-step explanation:

x/8 - 5 = -1

Add 5 both sides : x/8 = 4

Multiply both sides by four: x=4 times 8

4 0
3 years ago
Noelle ran 4 1/2 miles in the morning and 3 1/4 miles in the evening. How far did Noelle run that day?
KIM [24]
You just add all the numbers together:
3 1/8+ 4 1/2+ 4 1/4
=25/8+9/2+17/4
=25/8+36/8+34/8
=95/8
=11 7/8

Ugh.. did you type in the numbers correctly? My guess would be it was meant to be B tho :)

6 0
4 years ago
Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).
Oduvanchick [21]
For two complex numbers z_1=re^{i\theta}=r(\cos\theta+i\sin\theta) and z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi), the product is

z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))

That is, you multiply the moduli and add the arguments. You have z_1=7e^{i40^\circ} and z_2=6e^{i145^\circ}, so the product is

z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}
3 0
3 years ago
Read 2 more answers
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