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3 years ago

4/1=12/3,so 12 cups = quarts

Mathematics

2 answers:

REY [17]3 years ago

5 0

Yeah I think so I just took a wild guess and looked at my measurement sheet

m_a_m_a [10]3 years ago

5 0

The right answer is 12 cups = 3 quarts

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1. A state is considering adding another character to their vehicle license plates in order to increase the number of license pl

frez [133]

9514 1404 393

Answer:

a) 1,544,804,416

b) 52,523,350,144

c) 34

Step-by-step explanation:

(a) Each character position can be any of 24 letters or 10 digits, so any of 34 characters. The number of possible characters sets increases by a factor of 34 for each added character. A six-character plate will have ...

34^6 = 1,544,804,416 . . . possibilities

(b) 34^7 = 52,523,350,144 . . . possibilities

4 0

3 years ago

Use vertical multiplication to find the product of:

Rom4ik [11]

Answer:

$x^6+x^4+4x^3-2x^2-x+3$

Step-by-step explanation:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

First step multiply your terms in your first expression just to the 1 in the second expression like so:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$ Anything times 1 is that anything.

That is, $(x^3+2x+3) \cdot 1=x^3+2x+3$ .

Now we are going to take the top expression and multiply it to the -x in the second expression. $-x(x^3+2x+3)=-x^4-2x^2-3x$ . We are going to put this product right under our previous product.

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$

$-x^4-2x^2-3x$

We still have one more multiplication but before we do that I'm going to put some 0 place holders in and get my like terms lined up for the later addition:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^4+x^3+0x^2+2x+3$

$-x^4+0x^3-2x^2-3x+0$

Now for the last multiplication, we are going to take the top expression and multiply it to x^3 giving us $x^3(x^3+2x+3)=x^6+2x^4+3x^3$ . (I'm going to put this product underneath our other 2 products):

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^4+x^3+0x^2+2x+3$

$-x^4+0x^3-2x^2-3x+0$

$x^6+2x^4+3x^3$

I'm going to again insert some zero placeholders to help me line up my like terms for the addition.

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^6+0x^4+x^3+0x^2+2x+3$

$0x^6-x^4+0x^3-2x^2-3x+0$

$x^6+2x^4+3x^3+0x^2+0x+0$

----------------------------------------------------Adding the three products!

$x^6+x^4+4x^3-2x^2-x+3$

8 0

4 years ago

a cookie recipe calls for 2 cups of sugar and makes 36 cookies.If sarah only wants to make 24 cookies,how much sugar would she n

laiz [17]

Answer:

2=26

?=24

2×24

=1 11 cups of sugar

7 0

3 years ago

Please help me out thanks

Katyanochek1 [597]

Addition problems that need annex zeros:

Choice A. - 0.54 + 12.1

Choice C. - 25.59 + 1.2

Choice D. - 23.04 + 124.1 + 34.06

Hope I was able to help! :)

8 0

3 years ago

What are the roots of this equation 2x^2+7x-3=0

Ludmilka [50]

Step-by-step explanation:

2x^2+7x-3=0

by doing splitting the middle term we get/ using quadratic eq we get-

Let's solve your equation step-by-step.

2x2+7x−3=0

For this equation: a=2, b=7, c=-3

2x2+7x+−3=0

Step 1: Use quadratic formula with a=2, b=7, c=-3.

−b±√b2−4ac

−(7)±√(7)2−4(2)(−3)

2(2)

x= −7±√73 /4

x= −7 4 + 1 4 √73 or x= −7 4 + −1 4 √73

Answer:

x= −7 4 + 1 4 √73 or x= −7 4 + −1 4 √73

please mark me as the brainliest:)

8 0

4 years ago

Read 2 more answers