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inessss [21]
2 years ago
9

-2<3 multiple each side by -5 what is the resulting inequality

Mathematics
2 answers:
fgiga [73]2 years ago
8 0

Answer:

\boxed{10>-15}

Step-by-step explanation:

-2

Multiply each side by -5.

-2 \times -5

Switch signs when multiplying or dividing by a negative integer.

10>-15

sergeinik [125]2 years ago
8 0

Answer:

10 > -15

Step-by-step explanation:

-2 < 3

We want to multiply each side by -5 and we want to see what happens:

-2 * -5 < 3 * -5

= 10 < -15

This is a wrong inequality as 10 is ahead of -15 on the number line.

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You have to fill in the blank for this one 4/9 divided by blank equals 12
photoshop1234 [79]

For this case, the first thing we must do is define a variable.

We have then:

x: unknown number

We now write the expression that models the problem:

\frac{\frac{4}{9}}{x} = 12

From here, we clear the value of x.

We have then:

\frac{4}{9}=12x

\frac{4}{9(12)}=x

x=\frac{4}{108}

x=\frac{1}{27}

Answer:

4/9 divided by 1/27 equals 12

6 0
2 years ago
Read 2 more answers
If you received a score of 80% on an assignment,what is the ratio of correct answers to incorrect answers
Cerrena [4.2K]
Im not really sure so ill just give it a try.
mayeb 70? ten less"
6 0
2 years ago
Y - 4x = 5<br> 8x - 2y = 16<br> Solve using system of equations
kumpel [21]

Answer:

No solution

Step-by-step explanation:

Let's solve this system using elimination through addition/subtraction.  Multiply the second equation by (1/2), which results in the system

y - 4x = 5

4x - y = 8

--------------

0x + 0y = 13.  This is NEVER true.  This system of equation has NO  SOLUTION.

8 0
3 years ago
5x^2 -15x-140 what method should be use?
elena55 [62]
5x^2 -15x-140

5(x² -(3x) -28)

5(x² +(4x-7x)-28)

5(x(x+4) -7(x+4)

5(x-7)*(x-4)



8 0
2 years ago
Triangles ΔABC ≅ ΔBAD so that C and D lie in the opposite semi-planes of segment AB. Prove that segment CD bisects segment AD.
kolbaska11 [484]

Answer:

See explanation

Step-by-step explanation:

Triangles ΔABC and ΔBAD are congruent. So,

  • AB ≅ BA;
  • AC ≅ BD;
  • BC ≅ AD;
  • ∠ABC ≅ ∠BAD;
  • ∠BCA ≅ ∠ADB;
  • ∠CAB ≅ ∠DBA.

Consider triangles AEC and BED. In these triangles,

  • AC ≅ BD;
  • ∠EAC ≅ ∠EBD (because ∠CBA ≅ ∠BAD);
  • ∠AEC ≅ ∠BED (as vertical angles).

So, ΔAEC ≅ ΔBED. Thus,

AE ≅ EB.

This means that segment CD bisects segment AD.

6 0
3 years ago
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