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KengaRu [80]
2 years ago
7

Which inequalities would have a closed circle when graphed? check all that apply. x > 2.3 5.7 less-than-or-equal-to p one-hal

f greater-than y m greater-than-or-equal-to 10 s < –7.6
Mathematics
1 answer:
aalyn [17]2 years ago
6 0

Option B and D would have closed circles when graphed

The domain and range of a function are the components of a function. The domain is the set of all the input values of a function and range is the possible output given by the function.

At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality.

∴ Option B and D that are 5.7 ≤ p and \frac{1}{2} ≥ y respectively would have closed circles

Learn more about domain and range here :

brainly.com/question/13856645

#SPJ4

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3 0
3 years ago
Multiply the polynomials *<br> –6w(2w2 - 5)
Paul [167]

Answer:

-12w^3 + 30w.

Step-by-step explanation:

–6w(2w^2 - 5)        Using the distributive law:

= -6w* 2w^2 - 6w*-5

= -12w^3 + 30w.

5 0
4 years ago
A large bowl holds 5 2/3 cups of punch. Some more punch is added to the bowl to make a total of 10 cups of punch. Let represent
tatuchka [14]

Answer:

4 1/3 cups

Step-by-step explanation:

All you have to do for this one is subtract 5 2/3 from 10, which is 4 1/3. Your equation could be 10-5 2/3=x.

5 0
3 years ago
Ayuda con estos ejercicios
joja [24]
1. x=6
2. y=6
3. x=10
4. y=10

1.c+9= 19
2. 9-y=7
3. 5y=30
4. u-u=0

1. 2(3x2 - 5x3)
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3 0
3 years ago
Solve for x.
egoroff_w [7]

∑ Hey, KLPJDP615 ⊃

Answer:

x = 6 or x = -10

Step-by-step explanation:

<u><em>Given:</em></u>

<em>Solve for x.</em>

<em>1 + |2+x|= 9</em>

<em>O x = 4 or x = -8</em>

<em>O x = 7 or X = -11</em>

<em>O x = 5 or x = -9</em>

<em>x = 6 or X = -10 </em>

<u><em>Solve:</em></u>

<em>1 + |2+x|= 9</em>

<em>Subtract 1 from both sides:</em>

<em>1 + |2 +x| -1 = 9-1 </em>

<em>Simplify</em>

<em>|2 + x | = 8</em>

<em>Applying absolute value rule:  If |u| = a, a > 0 then u = a or u = -a</em>

<em>2 + x = -8</em>

<em>2 + x = 8</em>

<u><em>Solving:</em></u>

<em>2 + x = -8</em>

<em>2 - 2 + x = -8 - 2</em>

<em>x = -10</em>

<u><em>Solving:</em></u>

<em>2 + x = 8</em>

<em>2 - 2 + x = 8 - 2</em>

<em>x = 6</em>

<em />

<em>Hence, x = 6 or x = -10</em>

<em />

<u><em>xcookiex12</em></u>

<em>8/26/2022</em>

3 0
2 years ago
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