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3 years ago

15 quarts=_cups how many cups???

Mathematics

2 answers:

atroni [7]3 years ago

7 0

Answer:

60 cups

Step-by-step explanation:

Mazyrski [523]3 years ago

3 0

60 c cupssswssss

STEP
STEP
STEP

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What is a number between 7.833 and 7.834

dalvyx [7]

7.8335 is a number between 7.833 and 8.834

3 0

3 years ago

Read 2 more answers

Help asap check picture

Rudik [331]

The answer will be option D {(R,1),(R,2)(R,3)(R,4(R,5)(R,6),(B,1)(B,2)(B,3)(B,4)(B,5)(B,6).

<h3>What is probability?</h3>

The probability is defined as the chances of happening of any random event. or the possibility of happening of any event is called as the probability.

It is given in the question that there are two colours in the spinner and the dice have 6 faces from {1 to 6} numbers on it.

The sample outcomes with the spinner and the dice will be as follows:-

S= {(R,1),(R,2)(R,3)(R,4(R,5)(R,6),(B,1)(B,2)(B,3)(B,4)(B,5)(B,6).

Hence answer will be {(R,1),(R,2)(R,3)(R,4(R,5)(R,6),(B,1)(B,2)(B,3)(B,4)(B,5)(B,6).

To know more about probability follow

brainly.com/question/24756209

#SPJ1

5 0

2 years ago

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

pickupchik [31]

<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$ .

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$ .

$x^4-41x^2+400=0$ .

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

$(x^2-25)(x^2-16) =0$

Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

$(x^2-25) = x^2-5^2= (x-5)(x+5)$ and

$x^2-16 = x^2-4^2 =(x-4)(x+4)$ .

Therefore,

$(x-5)(x+5)(x-4)(x+4) =0$

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

6 0

2 years ago

20 POINTS PLEASE HELP MEEEE

Triss [41]

its actually 10 points

x^4 + 9x^3 + 15x^2 + 9x + 14

Split 15x^2 into 14x^2 + x^2

x^4 + 9x^3 + 14x^2 + x^2 + 9x + 14 factor as

x^2 [x^2 + 9x + 14 ] + 1 [ x^2 + 9x + 14 ] take out GCF

[ x^2 + 1 ] [ x^2 + 9x + 14 ]

[ x^2 + 1 ] [ x + 7 ] [ x + 2 ]

mark brainliest

6 0

3 years ago

Read 2 more answers

Please no link answers you will be reported

rewona [7]

Answer:

Step-by-step explanation:

You can use the key features (such as the vertex) to determine which is the correct equation

7 0

2 years ago