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deff fn [24]
3 years ago
5

True or False? The number 0.8 can be written as 8/10, so it is an irrational number

Mathematics
2 answers:
inysia [295]3 years ago
5 0
False it’s a rational number
lesya692 [45]3 years ago
3 0
The answer would be false because the number .8 can be 8/10 but it is rational and you can understand that. It would not be an irrational number.
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3x^2 + 2x + 5 = 0
Lyrx [107]

Answer:

Step-by-step explanation:

  3x2−2x−5=0

Let us factorize using the split the middle term method.

On inspection we observe that middle term can be split in two parts whose product is equal to the product of first and third term.

Two terms are −5xand3x

The equation becomes

3x2+3x−5x−5=0, pairing the two and taking out common factors

⇒(3x2+3x)−(5x+5)=0,

don't forget to change the sign of −5 once placed inside the parenthesis.

⇒3x(x+1)−5(x+1)=0

⇒(x+1)(3x−5)=0

Either (x+1)=0 or (3x−5)=0

We obtain x=−1,53

or x=−1,123

3 0
2 years ago
The approximate volume in milliliters, y, for a volume of x fluid ounces is equal to 29.57 times the value of x. Which table rep
Angelina_Jolie [31]

The table which represents the relationship is:

Fluid ounces, x            Volume y ( in millimeters)

       1                                       29.57

       2                                      59.14

       3                                      88.71

       4                                     188.28

In mathematics, a relation describes how two distinct sets of information related to one another. If more than two non-empty sets are taken into consideration, a connection between their elements will indicate that more than two sets are being evaluated.

Let y be the volume of fluid x in millimeters.

Now, y is equal to 29.57 times the value of x.

Therefore,

y = 29.57 × x

y = 29.75x

So, when x = 1:

y = 29.75(1) = 29.57

When x = 2,

y = 29.57(2) = 59.14

When x = 3,

y = 29.57(3) = 88.71

When x = 4,

y = 29.57(4) = 118.28

Therefore, the table that represents the relationship is:

Fluid ounces, x            Volume y ( in millimeters)

       1                                       29.57

       2                                      59.14

       3                                      88.71

       4                                     188.28

Learn more about volume here:

brainly.com/question/463363

#SPJ1

8 0
1 year ago
Find the values of the six trigonometric functions for angle θ.Give answers in simplest form.
natita [175]

Answer:

sin (theta) = \frac{\sqrt{15} }{8}

cos (theta) = \frac{7}{8}

tan (theta) = \frac{\sqrt{15} }{7}

cot (theta) = 7 square root 15/15

sec (theta) = \frac{8}{7}

csc (theta) = 8 square root 15/15

Step-by-step explanation:

7 0
3 years ago
Kiara bought fabric to make a dress and a sewing machine. The total cost of her purchase, $80.50, included both the cost of her
Mama L [17]
The cost is of fabric per foot 40.25$
4 0
2 years ago
In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2

Case k = −1:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

5 0
3 years ago
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