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Papessa [141]
3 years ago
7

1. Is 5(x+1) the same as 5x+1? Explain your

Mathematics
1 answer:
tatiyna3 years ago
5 0

Answer:

Step-by-step explanation:

5(x+1), when expanded, becomes 5x + 5, which is ot the same as 5x + 1.

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Will mark brainliest! IMPORTANT
enot [183]

Answer:

  P(x) = 3.88x -1750

Step-by-step explanation:

Profit is the difference between revenue and cost.

  P(x) = R(x) - C(x)

  P(x) = 4.1x -(0.22x +1750) . . . . . substitute the given expressions

  P(x) = 3.88x -1750 . . . . . . . . . . simplify

4 0
3 years ago
Yolanda is planning to put a pool in her backyard. The scaled model below gives the reduced measures for diameter and depth. The
atroni [7]

Answer:

C option: 6 feet

Step-by-step explanation:

Since Yolanda wants to keep the pool in proportion to the model, the ratio of diameter to depth of model and the pool will be same.

Let the depth of pool is x feet.

Ratio of Diameter to Depth of Model = Ratio of Diameter to Depth of pool

This means the depth of pool should be 6 feet if Yolanda wants to keep the pool in proportion to the model.

Plz give brainliest

4 0
3 years ago
Read 2 more answers
Simplify the following 3√28-5√63+4√112
alexandr402 [8]

7√7

using the ' rule of radicals '

• √a × √b ⇔ √ab

simplifying the radicals

√28 = √(4 × 7 ) = √4 × √7 = 2√7

√63 = √(9 × 7) =√9 × √7 = 3√7

√112 = √(16 × 7 ) = √16 × √7 = 4√7

substituting into the expression

3(2√7) - 5(3√7) + 4(4√7) = 6√7 - 15√7 + 16√7 = 7√7


7 0
3 years ago
Read 2 more answers
What is the relationship between segment ED and segment AC?
Bas_tet [7]
D midpoint of EC -----------------> FD parallel to AC and FD=AC/2=14/2=7 
<span>2-EB=EA E midpoint of AB </span>
<span>DB=DC D midpoint BC ...............> ED=AC/2=2 </span>
<span>3-T midpoint of SR </span>
<span>U midpoint of QR ---------> TU = QS/2 </span>
<span>QS=2 TU = 4.4 </span>
<span>4- The same steps SR=2 UV=9 </span>
<span>5-N midpoint of KM </span>
<span>O midpoint of ML </span>
<span>* NO parallel to Kl</span>
6 0
2 years ago
Could someone help me, please?
k0ka [10]

We can rewrite the expression under the radical as

\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4

then taking the fourth root, we get

\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|

Why the absolute value? It's for the same reason that

\sqrt{x^2}=|x|

since both (-x)^2 and x^2 return the same number x^2, and |x| captures both possibilities. From here, we have

\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|

The absolute values disappear on all but the b term because all of \dfrac32, a^2 and c^4 are positive, while b^3 could potentially be negative. So we end up with

\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4

3 0
3 years ago
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