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gayaneshka [121]
2 years ago
5

carl bought 0.8 pounds of peanuts and 0.3 pounds of raisins. how many pounds of snacks did he buy in all?

Mathematics
2 answers:
Readme [11.4K]2 years ago
7 0
0.3+0.8= 1.1
Carl brought 1.1 pounds of snacks.

Hope this helps!
alexandr402 [8]2 years ago
3 0

Answer:

Carla got 1.1 pounds of snacks.

Step-by-step explanation:

When Carla bought 0.8 pounds of peanuts and 0.3 pounds of raisins. To solve this problem, make sure to add 0.8+0.3, which give us equal to 1.1. 1.1 is the correct answer.

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PLZ HELP!!
ludmilkaskok [199]

9514 1404 393

Answer:

  y = -3x^2 -9x +30

Step-by-step explanation:

When a polynomial has a zero at x=p, it has a factor of (x -p). The factors of your quadratic are ...

  f(x) = (x +5)(x -2)

At the point x=3, the value of this product is ...

  f(3) = (3 +5)(3 -2) = (8)(1) = 8

In order for that value to be -24, it needs to be multiplied by a scale factor of -3. The quadratic you want is ...

  y = -3(x +5)(x -2)

  y = -3x^2 -9x +30 . . . . . standard form

3 0
3 years ago
Find the area of the circle r=14m use pi=3.14 and round your answer to the nearest hundredth
pantera1 [17]

Answer:

<h2>The answer is 615 m²</h2>

Step-by-step explanation:

Area of a circle = πr²

where

r is the radius

From the question

r = 14 m

π = 3.14

The area of the circle is

A = 3.14 × ( 14 ) ²

= 3.14 × 196

= 615.44

We have the final answer as

<h3>615 m² to the nearest whole number</h3>

Hope this helps you

7 0
3 years ago
team has a 75% chance to win each of the 3 games they will play this week. Clare simulates the week of games by putting 4 pieces
kaheart [24]

Answer: 88

Step-by-step explanation:

check google if its wrong :)

6 0
3 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
Help please live or death
iogann1982 [59]
Angle 2 is 143
angle 4 is 37
angle 7 is 143
angle 8 is 37
8 0
2 years ago
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