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algol [13]
2 years ago
5

What is the solution of the equation?  Eliminate any extraneous solutions.

Mathematics
1 answer:
vazorg [7]2 years ago
7 0
\sqrt{3x+28}-8=x\\\\\\A.\ x=9\ \vee\ x=-4\\x=9\to L=\sqrt{3\cdot9+28}-8=\sqrt{27+28}-8=\sqrt{55}-8\\ R=9;\\L\neq R\\\\B.\ x=-9\\L=\sqrt{3\cdot(-9)+28}-8=\sqrt{-27+28}-8=\sqrt1-8=1-8=-7;\\ R=-9;\ L\neq R

C.\ x=-9\ \vee\ x=-4\ (see\ B.)\\\\D.\ x=-4\\L=\sqrt{3\cdot(-4)+28}-8=\sqrt{-12+28}-8=\sqrt{16}-8\\=4-8=-4\\R=-4\\L=R\\\\Answer:D.
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3 \frac{1}{4}[/tex] = <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D" id="TexFormula1" title="\frac{1}{2}" alt="\frac{1
adelina 88 [10]

Answer:

Look below

Step-by-step explanation:

Conversion a mixed number 3 1/

4

to a improper fraction: 3 1/4 = 3 1/

4

= 3 · 4 + 1/

4

= 12 + 1/

4

= 13/

4

To find new numerator:

a) Multiply the whole number 3 by the denominator 4. Whole number 3 equally 3 * 4/

4

= 12/

4

b) Add the answer from previous step 12 to the numerator 1. New numerator is 12 + 1 = 13

c) Write a previous answer (new numerator 13) over the denominator 4.

Three and one quarter is thirteen quarters

Conversion a mixed number 2 1/

3

to a improper fraction: 2 1/3 = 2 1/

3

= 2 · 3 + 1/

3

= 6 + 1/

3

= 7/

3

To find new numerator:

a) Multiply the whole number 2 by the denominator 3. Whole number 2 equally 2 * 3/

3

= 6/

3

b) Add the answer from previous step 6 to the numerator 1. New numerator is 6 + 1 = 7

c) Write a previous answer (new numerator 7) over the denominator 3.

Two and one third is seven thirds

Add: 13/

4

+ 7/

3

= 13 · 3/

4 · 3

+ 7 · 4/

3 · 4

= 39/

12

+ 28/

12

= 39 + 28/

12

= 67/

12

For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(4, 3) = 12. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 4 × 3 = 12. In the next intermediate step, the fraction result cannot be further simplified by canceling.

In words - thirteen quarters plus seven thirds = sixty-seven twelfths.

3 0
2 years ago
What are the first two answers? #16 and #17
Luden [163]
Answer to number 16 is 27 and 17 is 2
5 0
3 years ago
One bottle contains 250 ml of juice. Soham wants 2 liters of juice. How many bottles should he buy?
Brrunno [24]

Answer:

8 bottles

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
What is the area of a sector with a central angle of 210 degrees and a diameter of 4.6m?
WINSTONCH [101]
Sector area = (central angle / 360) * PI * radius^2
sector area = (210 / 360) * PI * 2.3^2
sector area = (7 / 12) * PI * 5.29
<span><span><span>sector area = 9.6944313302 </span> </span> </span>
<span>sector area = 9.7 square meters (rounded)

Source:
http://www.1728.org/radians.htm

</span>
8 0
3 years ago
Given f(x)=e^-x^3 find the vertical and horizontal asymptotes
Aleonysh [2.5K]

Given:

f\mleft(x\mright)=e^{-x^3}

To find the vertical and horizontal asymptotes:

The line x=L is a vertical asymptote of the function f(x) if the limit of the function at this point is infinite.

But, here there is no such point.

Thus, the function f(x) doesn't have a vertical asymptote.

The line y=L is a vertical asymptote of the function f(x) if the limit of the function (either left or right side) at this point is finite.

\begin{gathered} y=\lim _{x\rightarrow\infty}e^{-x^3} \\ =e^{-\infty} \\ y=0 \\ y=\lim _{x\rightarrow-\infty}e^{-x^3} \\ y=e^{\infty} \\ =\infty \end{gathered}

Thus, y = 0 is the horizontal asymptote for the given function.

5 0
1 year ago
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