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vaieri [72.5K]
3 years ago
11

Which of the following is the correct graph of the linear equation below? y+2=1/5(x-1)

Mathematics
1 answer:
kupik [55]3 years ago
8 0

Answer:

Step-by-step explanation:

You might be interested in
16/x = 2/7 cross multiplication
WITCHER [35]

Answer:

x=56

Step-by-step explanation:

16 x 7= x * 2

x=\frac{16 x7}{2}

x=56

Hope this helps :)

8 0
3 years ago
What is the solution to the equation
Dafna11 [192]
\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6

therefore
D:x\in\left< -\dfrac{5}{4};\ 6\right>

\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs
9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D
Answer: t = -1.


7 0
3 years ago
Read 2 more answers
Calculate the annual interest for a $10,000 Treasury bond with a current yield of 3% that is quoted at 106 points.
Sergio [31]

Answer:

$318

Step-by-step explanation:

The treasury bond is $10,000

The current yield is 3%

= 3/100

=0.03

It is quoted at 106 points

The first step is to calculate the price of the bond

Price of the bond= $10,000×106/100

= $10,000×1.06

= $10,600

Therefore the annual interest can be calculated as follows

Annual interest= $10,600×0.03

= $318

Hence the annual interest is $318

7 0
3 years ago
What would be the answer?
iVinArrow [24]

Answer:

answer is C

Step-by-step explanation:

because -1<5

and not 6<-1 or -5>5

8 0
3 years ago
Help I’ll give brain crown
ankoles [38]

Answer:

13. c

14.a

15.c

16.d

17.b

18.d

19.a

20.b

Step-by-step explanation:

7 0
3 years ago
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