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nasty-shy [4]
3 years ago
8

Write an equation that describes the function.

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
4 0

Answer:

y = xplus1 sorry, new keyboard and it doesn't have a plus sign

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A. 5y2 - 60 = 0<br>by Square root property​
TEA [102]

Answer:

y  = ± 2sqrt(3)

Step-by-step explanation:

5y^2 - 60 = 0

Add 60 to each side

5y^2  = 60

Divide by 5

y^2 = 60/5

y^2 = 12

Take the square root of each side

sqrt( y^2) = ± sqrt(12)

y = ± sqrt(4*3)

y =  ±sqrt(4) sqrt(3)

y  = ± 2sqrt(3)

7 0
3 years ago
What is the value of n? 4x28+6x17-15=n
ad-work [718]

4 \times 28 = 112 + 6 \times 17 = 214 - 15 = 199
8 0
3 years ago
Read 2 more answers
Which of the following is the value of a when the function f(x) = 3|x| is written in the standard form of an absolute value
Gnoma [55]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Modulus functions.

Since here, the value of h and k are zero,hence

the value of the function f(x) = 3|x| is 3

====> answer is 3

4 0
3 years ago
Two faces of a six-sided die are painted red, two are painted blue, and two are painted yellow. The die is rolled three times, a
Keith_Richards [23]

Answer:

(a) 4/9

(b) 19/27

Step-by-step explanation:

The probability of each die turning up red or not turning red are:

P(R) =\frac{2}{6}= \frac{1}{3}\\P(O) =\frac{4}{6}= \frac{2}{3}

(a). Exactly one face up is red.

The odds of only the first die being red are:

P(A=R) =\frac{1}{3} *\frac{2}{3}* \frac{2}{3} \\P(1=R) =\frac{4}{27}

The same odds are valid for only the second or only the third die being red. Therefore, the probability that exactly one die turns up red is:

P(R=1) = 3*\frac{4}{27}\\P(R=1) = \frac{4}{9}

(b). At least one face up is red.

The probability that at least one die turns up red is the sum of the probabilities of exactly one (found in the previous item), two or all dice turning up red.

Following the same logic as in the previous item, the probability that exactly two dice turn up red is:

P(R=2) = 3*(\frac{1}{3}*\frac{1}{3}*\frac{2}{3})\\P(R=2) = \frac{2}{9}

The probability that all die turn up red is:

P(R=3) = \frac{1}{3}*\frac{1}{3}*\frac{1}{3}\\P(R=3)=\frac{1}{27}

Thus, the probability that at least one die turns up red is:

P= P(R=1)+P(R=2)+P(R=3) = \frac{4}{9}+\frac{2}{9}+\frac{1}{27}\\P=\frac{19}{27}

4 0
3 years ago
Jimmy set his score for his math test. After the math teacher returnee his paper to him, he realized that if he increased his ta
mixer [17]
When ever you have percentages, it should be helpful to bear in mind you can express them as multipliers. In this case, it will be helpful.
So, if we let:
a = test score
b = target score
then, using the information given:
a = 1.1b + 1
a = 1.15b - 3
and we get simultaneous equations.
'1.1' and '1.15' are the multipliers that I got using the percentages. Multiplying a value by 1.1 is the equivalent of increasing the value by 10%. If you multiplied it by 0.1 (which is the same as dividing by 10), you would get just 10% of the value.
Back to the simultaneous equations, we can just solve them now:
There are a number of ways to do this but I will use my preferred method:
Rearrange to express in terms of b:
a = 1.1b + 1
then b = (a - 1)/1.1
a = 1.15b - 3
then b = (a + 3)/1.15
Since they are both equal to b, they are of the same value so we can set them equal to each other and solve for a:
(a - 1)/1.1 = (a + 3)/1.15
1.15 * (a - 1) = 1.1 * (a + 3)
1.15a - 1.15 = 1.1a + 3.3
0.05a = 4.45
a = 89

8 0
3 years ago
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