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ivolga24 [154]
3 years ago
6

(1 pt) Which graphs show the solution to the given inequalities? Column A Column B 1. 2. 3. 4. A. Number line with positive and

negative numbers. A ray starts at negative 12 with an open circle and extends to the right. B. Number line with positive and negative numbers. A ray starts at negative 12 with a solid circle and extends to the right. C. Number line with positive and negative numbers. A ray starts at negative 12 with a solid circle and extends to the left. D. Number line with positive and negative numbers. A ray starts at negative 12 with an open circle and extends to the left.
Mathematics
1 answer:
ladessa [460]3 years ago
4 0

I don't know but its at the tip of my tounge


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Efectuati calculele:
NNADVOKAT [17]
\sqrt{(5 - 1)^{2}} +  \sqrt{20}

\sqrt{4^{2}} +  \sqrt{4 * 5}

\sqrt{16} +  \sqrt{4} \sqrt{5} 


4 + 2 \sqrt{5} 


4\sqrt{5} + 2 \sqrt{5} 


6 \sqrt{5}


\sqrt{(10 - 3)^{2}} +  \sqrt{20 * 18}

\sqrt{7^{2}}  +  \sqrt{4 * 9 * 5 * 2}

\sqrt{49} +  \sqrt{36 * 10}

7 +  \sqrt{36}\sqrt{10}

7 + 6\sqrt{10}

7\sqrt{10} + 6\sqrt{10}

13\sqrt{10}


\sqrt[3]{(3 + 1)^{2}} -  \sqrt{(3 - 2)x^{2}

\sqrt[3]{4^{2}} -  \sqrt{x^{2}

\sqrt[3]{16} - x[tex] \\[tex] \sqrt[3]{8 * 2} - x

\sqrt[3]{8} \sqrt[3]{2} - x


2\sqrt[3]{2} - x




8 0
3 years ago
I need helppppopppppppppp
Ivan

Answer:

we all do were all gonna die one day

Step-by-step explanation:

6 0
2 years ago
How would the expression x3 +8 be rewritten using Sum of Cubes?
Ipatiy [6.2K]

Answer:

alr so ques is

x3 + 8

x^3 + 2^3

now apply identity of a3 + b3

so finally we get

=(x+2)(x2−2x+4)

option 2

enjoy!!!!!

8 0
3 years ago
Find the value of x in each of the following<br>a) 5x/2+ 1 = 11<br>b) 2x/7 – 3 = 2​
laila [671]

Answer:

A) x=4

B) x=35/2

Step-by-step explanation:

calculate the product

A) 5x/2+1=11

multiply both sides of the equation by 2

5x+2=22

move the constant to the right hand side and change its sign

5x+22-2

now subtract the numbers

5x=20

now didvide both sides

5x=20

x=4

B)

calculate the product

2x/7-3=2

multiply both sides of the equation by 7

2x-21=14

move the constant to the right hand side and change its sign

2x=14+21

add the numbers

2x=35

now divide both sides of the equation by 2

x=35/2

8 0
3 years ago
The annual energy consumption of the town where Camilla lives in creases at a rate that is onal at any time to the energy consum
boyakko [2]

Answer:

6.575 trillion BTUs

Step-by-step explanation:

<em>Let represent the annual energy consumption of the town as E</em>

<em>The rate of annual energy consumption *  energy consumption at time past</em>

<em>dE/dt * E</em>

<em>dE/dt =K</em>

<em>k = the proportionality constant</em>

<em>c= the integration constant</em>

<em>(dE/dt=) kdt</em>

<em>lnE = kt + c</em>

<em>E(t) = e^kt+c ⇒ e^c e^kt  e^c is a constant, and e^c = E₀</em>

<em>E(t) = E₀ e^kt</em>

<em>The initial consumption of energy is E(0)=4.4TBTU</em>

<em>set t = 0 then</em>

<em>4.4 = E₀ e⇒ E₀ (1) </em>

<em>E₀ = 4.4</em>

<em>E (t) = 4.4e^kt</em>

<em>The consumption after 5 years is t = 5, e(5) = 5.5TBTU</em>

<em>so,</em>

<em>E(5) = 5.5 = 4.4e^k(5)</em>

<em>e^5k = 5/4</em>

<em>We now take the log 5kln = ln(5/4)</em>

<em>5k(1) = ln(5/4)</em>

<em>k = 1/5 ln(5/4) = 0.04463</em>

<em>We find  the town's annual energy consumption, after 9 years</em>

<em>we set t=9  </em>

<em>E(9) = 4.4e^0.04463(9)</em>

<em>= 4.4(1.494301) = 6.5749TBTUs</em>

<em>Therefore the annual energy consumption of the town after 9 years is </em>

<em>= 6.575 trillion BTUs</em>

<em />

3 0
3 years ago
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