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3 years ago

I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT.

Mathematics

2 answers:

Vikki [24]3 years ago

5 0

It is HE. Hope this helps!!!!! #Batgirl

Zepler [3.9K]3 years ago

5 0

FG. If you have any questions, message me.

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Worth 48 points!!!! plz help me.

Mrac [35]

Answer:

see below

Step-by-step explanation:

1. The inverse means take the not of both sides

If it is Saturday, then there is no school.

Inverse If it is not Saturday, then there is school.

The converse flips the then and the if

If there is no school, then it is Saturday.

2.) The first given part is the if, what will happen is the then part

If you are a freshman, then you are required to attend orientation.

3.)

Given: 3x+1=-14 Prove: x=-5

Statements Reasons

1. 3x+1=-14 Given

2. 3x=-15 Subtraction Property of equality ( since we subtract 1 from each side)

3. X=-5 Division Property ( since we divide each side by 3)

3 0

3 years ago

Read 2 more answers

Solve for x 11x+4<15 OR 12x−7>−25

never [62]

Isolate the x in both cases. What you do to one side, you do to the other.

11x + 4 < 15

Subtract 4 from both sides

11x + 4 (-4) < 15 (-4)

11x < 11

isolate the x, divide 11 from both sides

11x/11 < 11/11

x< 1

x < 1 is your answer for the first one

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Again, isolate the x.

12x - 7 > -25

Add 7 to both sides

12x - 7 (+7) > -25 (+7)

12x > -18

Isolate the x, divide 12 from both sides

12x/12 > -18/12

x > -1.5 is your answer for the second one.

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hope this helps

3 0

3 years ago

Read 2 more answers

I want the answer and way to do this problem.

user100 [1]

∠ABC = ∠CDA (given)

∠BCA = ∠DAC (given)

CA = AC (common side)

ASA congruence criterion = when two angles of two triangles are equal and their included side is also equal, we can consider those triangles to be congruent to each other.

Since two angles in triangle △ABC and△CDA are equal and since their included side is also equal we can conclude that they are congruent.

Therefore, △ABC ≅ △CDA under the ASA congruence criterion.

8 0

3 years ago

A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0

3 years ago

A technique that allows shoppers to compare prices?

Roman55 [17]

A) comparison shopping is the answer

3 0

3 years ago