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ki77a [65]
3 years ago
10

15) What is the value of y? (4y - 9) 40 5yº

Mathematics
1 answer:
love history [14]3 years ago
6 0

angle1 = 4y - 9°

angle2 = 40°

angle3 = 5y°

Angle sum property

180° = angle1 + angle2 + angle3

Therefore,

180 = (4y - 9) + 40 + 5y

180 = 4y - 9 + 40 + 5y

180 = 9y + 31

180 - 31 = 9y

149 = 9y

149/9 = y

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Morgarella [4.7K]

Answer:

55,784 oranges

Step-by-step explanation:

To fill the vats, they would need the number of oranges needed to fill one vat multiplied by the amount of vats they have.

734*76=55784

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3 years ago
Suppose that the functions q and r are defined as follows.
satela [25.4K]

Answer:

(r o g)(2) = 4

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Step-by-step explanation:

Given

g(x) = x^2 + 5

r(x) = \sqrt{x + 7}

Solving (a): (r o q)(2)

In function:

(r o g)(x) = r(g(x))

So, first we calculate g(2)

g(x) = x^2 + 5

g(2) = 2^2 + 5

g(2) = 4 + 5

g(2) = 9

Next, we calculate r(g(2))

Substitute 9 for g(2)in r(g(2))

r(q(2)) = r(9)

This gives:

r(x) = \sqrt{x + 7}

r(9) = \sqrt{9 +7{

r(9) = \sqrt{16}{

r(9) = 4

Hence:

(r o g)(2) = 4

Solving (b): (q o r)(2)

So, first we calculate r(2)

r(x) = \sqrt{x + 7}

r(2) = \sqrt{2 + 7}

r(2) = \sqrt{9}

r(2) = 3

Next, we calculate g(r(2))

Substitute 3 for r(2)in g(r(2))

g(r(2)) = g(3)

g(x) = x^2 + 5

g(3) = 3^2 + 5

g(3) = 9 + 5

g(3) = 14

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(q o r)(2) = 14

8 0
3 years ago
The average summer temperature in Anchorage is 69°F. If the daily temperature is normally distributed with a standard deviation
mafiozo [28]

Answer:

81.85%

Step-by-step explanation:

Given :

The average summer temperature in Anchorage is 69°F.

The daily temperature is normally distributed with a standard deviation of 7°F .

To Find:What percentage of the time would the temperature be between 55°F and 76°F?

Solution:

Mean = \mu = 69

Standard deviation = \sigma = 7

Formula : z=\frac{x-\mu}{\sigma}

Now At x = 55

z=\frac{55-69}{7}

z=-2

At x = 76

z=\frac{76-69}{7}

z=1

Now to find P(55<z<76)

P(2<z<-1)=P(z<2)-P(z>-1)

Using z table :

P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185

Now percentage of the time would the temperature be between 55°F and 76°F = 0.8185 \times 100 = 81.85\%

Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.

7 0
3 years ago
A solution to a system of equations is at the point of intersection.
solniwko [45]

Answer:

true

Step-by-step explanation:

i dont have an explanation its just true

7 0
3 years ago
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