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4 years ago

Helppppppppppppppppppppp

Mathematics

2 answers:

ivolga24 [154]4 years ago

7 0

Answer:

Step-by-step explanation:

Well, we can multiply the choices that they gave us, which is 3 by 4 by 3 by 3 by 10, which will give 1080 for the first part. Now, if you was to do only 3 toppings, it should be 3 by 4 by 3 by 3 by 3, which would give us 324. I don't know if this is right but I think this is the right answer.

egoroff_w [7]4 years ago

4 0

Answer:

a, c

Step-by-step explanation:

Let's start with one of each choice. That'll be 3 * 4 * 3 * 3 * 10 = 1080 pizzas. As for the second one, this time the toppings won't be 10 it will be 10 * 9 * 8 because we have 10, and then 9, and then 8 topping choices. This gives us 1080 * 9 * 8 = 77,760. Hope this helps!

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Question 3 of 10 What is the solution to this equation? 4(x-8) + 10 = -10 O A. x= 3 OB. x = 5 O c. x=2 O D. x= 4

krek1111 [17]

Answer:

O A. x = 3

Step-by-step explanation:

4(x - 8) + 10 = -10

4x - 32 + 10 = -10

4x = 12

x = 3

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3 years ago

Printer A can print 30 pages per minute. Printer B can print 40 pages per minute. Printer B begins printing 3 minutes after Prin

anzhelika [568]

Printer A prints for 10 minutes, so prints

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3 years ago

If (x+2) is a Factor x^3 + 2x^2 + 2x + k then find the value of K.

Delvig [45]

Answer:

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

-1.84 - 2.3m= -2.3(m+.8)

alexira [117]

Answer:

m ∈ R

Step-by-step explanation:

- 1,84 - 2,3m = - 2,3(m + 0,8)

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5 0

3 years ago

I really really really need help!!!!

Yakvenalex [24]

For $f$ to be continuous at $x=1$ , you need to have the limit from either side as $x\to1$ to be the same.

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b$

If $a=2$ and $b=3$ , then the limit from the right would be $2+3=5\neq2$ , so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation $a+b=2$ .

Part (c) is done similarly to part (1). This time, you need to limits from either side as $x\to2$ to match. You have

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b$
$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0$

So, $a$ and $b$ have to satisfy the relation $4a+2b=0$ , or $2a+b=0$ .

Part (4) is done by solving the system of equations above for $a$ and $b$ . I'll leave that to you, as well as part (5) since that's just drawing your findings.

8 0

3 years ago