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Otrada [13]
3 years ago
7

-0.9+28 ---- 6 help me solve the task im a little lazy ​

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
7 0

Answer:

around 4.52.

Fraction form: 4\frac{52}{100}

Step-by-step explanation:

Brainliest?

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Solve the Inequality: m+16>8m+2​
Setler [38]

Answer:

m<2

Step-by-step explanation:

To get to this, you first need to move the variable to one side, so you could subtract m from both sides to make the equation 16>7m+2. Next, you subtract 2 from both sides to make the equation into 14>7m. Finally, you would divide 7 from both sides to get 2>m, or m<2.

Hope this helps!

5 0
3 years ago
How much more area does a large pizza with a 12 in. diameter have than a small pizza with an 8 in. diameter? Round your answer t
ivolga24 [154]

Answer: About 63 in²

Step-by-step explanation:

<u>Area of circle = π · r²</u>

  • r = radius length
  • π ≈ 3.14

<u>Area of large pizza:</u>

\pi *r^{2} =3.14*6^{2} =3.14*36=113.04

<u>Area of small pizza:</u>

\pi *r^{2} =3.14*4^{2} =3.14*16=50.24

<u>Difference in area:</u>

113.04-50.24=62.8

4 0
3 years ago
Si Adrián tiene “x” años de edad, Teresa la tercera parte de Adrián, Isabel el doble de Teresa y Felipe el cuádruple de Adrián,
bixtya [17]

the answer is ISABELLA YOUR BOYFRINDS HERE

4 0
2 years ago
Find the central angle measure
Mariana [72]

Answer:

76

Step-by-step explanation

The angle across from 104 would be the same Leaving 152 left of 360 and 152 divided by 2 is 76

7 0
3 years ago
If p is a polynomial show that lim x→ap(x)=p(a
Lostsunrise [7]

Let p(x) be a polynomial, and suppose that a is any real number. Prove that

lim x→a p(x) = p(a) .

 

Solution. Notice that

 

2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .

 

So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x – 2.

 

Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

 

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

= 2|x|^3 + 5|x|^2 + |x| + 2

< 2(2)^3 + 5(2)^2 + (2) + 2

= 40

 

Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2 + x − 2| < ε/40 · 40 = ε.

3 0
3 years ago
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