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DanielleElmas [232]
1 year ago
12

Suppose you have developed a scale that indicates the brightness of sunlight. Each category in the table is 4 times brighter tha

n the next lower category. For example, a day that is dazzling is 4 times brighter than a day that is radiant. How many times brighter is a radiant day than a dim day?
Mathematics
1 answer:
raketka [301]1 year ago
8 0

According to the developed scale, a radiant day is <u>16 times</u> brighter than a dim day.

We assume the brightness of a dim day to be x.

According to the developed scale, the brightness of an illuminated day will be 4 times that of a dim day.

Thus, the brightness of an illuminated day = 4*the brightness of a dim day = 4x.

According to the developed scale, the brightness of a radiant day will be 4 times that of an illuminated day.

Thus, the brightness of a radiant day = 4*the brightness of an illuminated day = 4*4x = 16x.

Now, the ratio of the brightness of a radiant day to the brightness of a dim day = 16x:x = 16x/x = 16:1.

Thus, according to the developed scale, a radiant day is <u>16 times</u> brighter than a dim day.

Learn more about the developed scale at

brainly.com/question/4970963

#SPJ1

The question provided is incomplete. The complete question is:

"Suppose you have developed a scale that indicates the brightness of sunlight. Each category in the table is 4 times brighter than the next lower category. For example, a day that is dazzling is 4 times brighter than a day that is radiant. How many times brighter is a radiant day than a dim day?

Dim=2

Illuminated=3

Radiant=4

Dazzling=5"

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If a polynomial function f(x) has roots 6 and square root of 5, what must also be a root of f(x)?
ivolga24 [154]

Answer:

-\sqrt{5}

Step-by-step explanation:

A root with square root or under root is only obtained when we take the square root of both sides. Remember that when we take a square root, there are two possible answers:

  • One answer with positive square root
  • One answer with negative square root

For example, for the equation:

x^{2}=3

If we take the square root of both sides, the answers will be:

x=\sqrt{3} \text{ or } x= -\sqrt{3}

Only getting one solution with square root is not possible. Solutions with square root always occur in pairs.

For given case, the roots are 6 and \sqrt{5}. Therefore, the 3rd root of the polynomial function f(x) had to be -\sqrt{5}

It seems you made error while writing option B, it should be - square root of 5.

4 0
3 years ago
What is this number in standard form? (5×1,000)+(9×1)+(2×1/10)+(6×1/1,000)
mr Goodwill [35]

Answer:

5009.206

Step-by-step explanation:

<u>The answer is </u><em><u>5009.206</u></em>. We see that the first portion of this question asks, "(5×1,000)+(9×1)". 5x1000 is 5000. 9x1 is 9. 5000+9= 5009. The second portion of this question asks for the numbers that go after the decimal. 2x1/10 is 2/10. 6x1/1000 is 6/1000. The decimal place stands for one. So we do .206. If the decimal stands for ones that means in this case the 2 is in the tenths place (like it should be since the question asks for 2/10) and the 6 is in the 1000 place (like it should be since the question asks for 6/1000). We add this up together and our answer is 5009.206. Hope this explanation helps! Happy learning!

5 0
3 years ago
If f(p) divided by x-p and x-q have the same remainder
oee [108]
Hello,


x^2-y^2=(x+y)(x-y)
x^3-y^3=(x-y)(x²+xy+y²)

Let's use Horner's division

.........|a^3|a^2.|a^1..........|a^0
.........|1....|5....|6..............|8....
a=p...|......|p....|5p+p^2....|6p+5p^2+p^3
----------------------------------------------------------
.........|1....|5+p|6+5p+p^2|8+6p+5p^2+p^3

The remainder is 8+6p+5p^2+p^3 or 8+6q+5q^2+q^3


Thus:
8+6p+5p^2+p^3 = 8+6q+5q^2+q^3
==>p^3-q^3+5p^2-5q^2+6p-6p=0
==>(p-q)(p²+pq+q²)+5(p-q)(p+q)+6(p-q)=0
==>(p-q)[p²+pq+q²+5p+5q+6]=0 or p≠q
==>p²+pq+q²+5p+5q+6=0

And here, Mehek are there sufficients explanations?
3 0
2 years ago
3 consecutive integers for 273
a_sh-v [17]

Answer:

90, 91 and 92

Step-by-step explanation:

Given

Consecutive integers = 273

Required

Find the integers

The question seem to be incomplete. However, I'll assume we're dealing with sum.

Let the smallest integer be y.

So,

y + y + 1 + y + 2 = 273

Collect like terms.

y + y + y = 273 - 2 - 1

3y = 270

Divide both sides by 3

y = 90

Hence, the integers are 90, 91 and 92

8 0
2 years ago
How would you solve 7 - 3y = 22
nydimaria [60]
7-3y=22 -7 -7 -3y=15 Divide by -3. y=-5
7 0
3 years ago
Read 2 more answers
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