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2 years ago

I WILL GIVE BRAINIEST TO THE BEST ANSWER

Mathematics

2 answers:

Natalka [10]2 years ago

6 0

Domain is the values of x. x is from -5 to 7, it includes -5 ([ ] is the include symbol, but doesn't include 7, so C is the answer

djverab [1.8K]2 years ago

4 0

Answer:

Domain of the function given in picture is:

[-5,7)

Step-by-step explanation:

The domain of the function f(x)=y is all the possible values of x for which f(x) gives some value.

As we can see f(x) gives a table value for all the values of x between -5 and 7

where -5 is included but 7 is not included

Hence, domain of the function given in picture is:

[-5,7)

You might be interested in

Prove divisibility 45^45·15^15 by 75^30

Anuta_ua [19.1K]

Answer:

$3^{75}$

Step-by-step explanation:

We are asked to divide our given fraction: $\frac{45^{45}*15^{15}}{75^{30}}$ .

We will simplify our division problem using rules of exponents.

Using product rule of exponents $(a*b)^n=a^n*b^n$ we can write:

$45^{45}=(3*15)^{45}=3^{45}*15^{45}$

$75^{30}=(5*15)^{30}=5^{30}*15^{30}$

Substituting these values in our division problem we will get,

$\frac{3^{45}*15^{45}*15^{15}}{5^{30}*15^{30}}$

Using power rule of exponents $a^m*a^n=a^{m+n}$ we will get,

$\frac{3^{45}*15^{45+15}}{5^{30}*15^{30}}$

$\frac{3^{45}*15^{60}}{5^{30}*15^{30}}$

Using quotient rule of exponent $\frac{a^m}{a^n}=a^{m-n}$ we will get,

$\frac{3^{45}*15^{60-30}}{5^{30}}$

$\frac{3^{45}*15^{30}}{5^{30}}$

Using product rule of exponents $(a*b)^n=a^n*b^n$ we will get,

$\frac{3^{45}*(3*5)^{30}}{5^{30}}$

$\frac{3^{45}*3^{30}*5^{30}}{5^{30}}$

Upon canceling out $5^{30}$ we will get,

$3^{45}*3^{30}$

Using power rule of exponents $a^m*a^n=a^{m+n}$ we will get,

$3^{45+30}$

$3^{75}$

Therefore, our resulting quotient will be $3^{75}$ .

8 0

2 years ago

Help Me Number 6 or if you can help me with number 7 if you want too ☺️

Makovka662 [10]

Simply use the formula they give you. For question 6, 4 is the radius. So the area of the circle is 4pi. Now for seven you do the same thing, but they gave you the diameter instead and it’s a half circle. So try that one on your own. Let me know if you get stuck

7 0

3 years ago

The expression -6 -7(4+3x) is equivent to

Ipatiy [6.2K]

Answer:

Done!

Step-by-step explanation:

-6 - 7(4 + 3x)

Distribute;

-6 - 28 - 21x

-34 - 21x

5 0

2 years ago

15.30 find the inverse laplace transform of: 1. (a) f1(s) = 6s 2 8s 3 s(s 2 2s 5) 2. (b) f2(s) = s 2 5s 6 (s 1) 2 (s 4) 3. (c) f

EleoNora [17]

The solution of the inverse Laplace transforms is mathematically given as

$f_{1}(t)=e^{-t}\sin (2 t)$

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$

<h3>What is the inverse Laplace transform?</h3>

Generally, the equation for the function is mathematically given as

$F_{1}(s)=\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}$

By Applying the Partial fractions method

$\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}=\frac{A}{s}+\frac{B s+C}{s^{2}+2 s+5}$

$6 s^{2}+8 s+3=A\left(s^{2}+2 s+5\right)+(B s+C) s$

$\begin{aligned}&3=5 A \\&A=\frac{3}{5}\end{aligned}$

Considers s^2 coefficient

$\begin{aligned}&6=A+B \\&B=6 \cdot A \\&B=\frac{27}{5}\end{aligned}$

Consider s coeffici ent

$\begin{aligned}&8=2 A+C \\&C=8-2 A \\&C=\frac{34}{5}\end{aligned}$

Putting these values into the previous equation

$&F_{1}(s)=\frac{3}{5 s}+\frac{27 s+34}{5\left(s^{2}+2 s+5\right)} \\\\&F_{1}(s)=\frac{3}{5 s}+\frac{27(s+1)}{5\left((s+1)^{2}+4\right)}+\frac{7 \times 2}{10\left((s+1)^{2}+4\right)}$

By taking Inverse Laplace Transforms

$f_{1}(t)=\frac{3}{5}+\frac{27}{5} e^{-t} \cos (2t) + \frac{7}{10}\\\\$

$f_{1}(t)=e^{-t}\sin (2 t)$

For B

$F_{2}(s)=\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}$

By Applying Partial fractions method

$\begin{aligned}&\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\\\&s^{2}+5 s+6=A(s+1)(s+4)+B(s+4)+C(s+1)^{2}\end{aligned}$

at s=-1

$1-5+6=3 B \\\\B=\frac{2}{3}$

at s=-4

$&16-20+6=9 C \\\\&9 C=2 \\\\&C=\frac{2}{9}$

at s^2 coefficient

1=A+C

A=1-C

A=7/9

inputting Variables into the Previous Equation

$\begin{aligned}&F_{2}(s)=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\&F_{2}(s)=\frac{7}{9(s+1)}+\frac{2}{3(s+1)^{2}}+\frac{2}{9(s+4)}\end{aligned}$

By taking Inverse Laplace Transforms

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

For C

$F_{3}(s)=\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}$

Using the strategy of Partial Fractions

$\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}=\frac{A}{s+1}+\frac{B s+C}{s^{2}+4 s+8}$

$10=A\left(s^{2}+4 s+8\right)+(B s+C)(s+1)$

S=-1

10=(1-4+8) A

A=10/5

A=2

Consider constants

10=8 A+C

C=10-8 A

C=10-16

C=-6

Considers s^2 coefficient

0=A+B

B=-A

B=-2

inputting Variables into the Previous Equation

$&F_{3}(s)=\frac{2}{s+1}+\frac{-2 s-6}{\left((s+2)^{2}+4\right)} \\\\&F_{3}(s)=\frac{2}{s+1}-\frac{2(s+2)}{\left((s+2)^{2}+4\right)}-\frac{2}{\left((s+2)^{2}+4\right)}$

Inverse Laplace Transforms

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$