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3 years ago

Enter a negative number that is greater than -10. btw this is iready

Mathematics

2 answers:

Molodets [167]3 years ago

6 0

Answer:

5 6 7 8 9 10 All the positive number are grater than -10 but if it askes for negative you would you -5 -6 -7 -8 -9

Step-by-step explanation:

konstantin123 [22]3 years ago

6 0

With all numbers, whether they are negative, zero or positive: A number is greater than another when it is to the right on the number line. Conversely a number is less than another when it is to the left on the number line.

Step-by-step explanation:

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Can someone explain how to do this please?<br><br> 3 I2x-3I+2=3

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3 years ago

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3 years ago

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2 years ago

7.3 homework help me

olga_2 [115]

Answer:

1. Yes

∆RST ~ ∆WSX

by SAS

2. Yes

∆ABC ~ ∆PQR

by SSS

3. Yes

∆STU ~ ∆JPM

by SAS

4. Yes

∆DJK ~ ∆PZR

by SAS

5. Yes

∆RTU ~ ∆STL

by SAS

5. Yes

∆JKL ~ ∆XYW

by SAS

6. No

7. Yes

∆BEF ~ ∆NML

by SAS

8. Yes

∆GHI ~ ∆QRS

by SSS

9. x=22

10. x=12

Step-by-step explanation:

1. RS/WS=ST/SX and m<RST=m<WSX

2. AB/PQ=8/6=4/3

BC/QR=AC/PR=12/9=4/3

AB/PQ=BC/QR=AC/PR

3. ST/JP=10/15=2/3

SU/JM=14/21=2/3

ST/JP=2/3=SU/JM

and m<TSU=70°=m<PJM

4. DK/PR=8/4=2

JK/ZR=18/9=2

DK/PR=2=JK/ZR

and m<DKJ=65°=m<PRZ

5. RT/ST=UT/LT

and m<RTU=m<STL

6. KL/YW=20/18=10/9

JL/XW=36/24=3/2

KL/YW=10/9≠3/2=JL/XW

7. BF/NL=24/16=3/2

BE/NM=39/26=3/2

BF/NL=3/2=BE/NM

and m<EBF=m<MNL

8. GH/QR=32/20=8/5

HI/RS=40/25=8/5

GI/QS=24/15=8/5

GH/QR=HI/RS=GI/QS=8/5

9. x/33=18/27

Simplifying the fraction on the right side of the equation:

x/33=2/3

Solving for x: Multiplying both sides of the equation by 33:

33(x/33)=33(2/3)

x=11(2)

x=22

10. x/16=9/12

Simplifying the fraction on the right side of the equation:

x/16=3/4

Solving for x: Multiplying both sides of the equation by 16:

16(x/16)=16(3/4)

x=4(3)

x=12

4 0

3 years ago

A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v>=2i-4tj^

guapka [62]

Given :

A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj .

To Find :

A. The vector position of the particle at any time t .

B. The acceleration of the particle at any time t .

Solution :

A )

Position of vector v is given by :

$d=\int\limits {v} \, dt\\\\d=\int\limits {(2i-4tj)} \, dt \\\\d=(2t)i+\dfrac{4t^2}{2}j\\\\d=(2t)i+(2t^2)j$

B )

Acceleration a is given by :

$a=\dfrac{dv}{dt}\\\\a=\dfrac{2i-4tj}{dt}\\\\a=\dfrac{2i}{dt}-\dfrac{4tj}{dt}\\\\a=0-4j\\\\a=-4j$

Hence , this is the required solution .

5 0

3 years ago