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Liula [17]
3 years ago
5

Can i get help plzz .....

Mathematics
1 answer:
wel3 years ago
3 0

Answer:

$0.30

Step-by-step explanation:

If you already spent $5 then you have $6 remaining so you will just divide 6 and 20 and you will get 0.3 so $0.30

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- 65 POINTS -
eduard

Answer: f(x) = ∛(x+4)^2=8 + 3 = 11

4x5=20-9=11

Step-by-step explanation:

3 0
3 years ago
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I’m on khan academy(6th grade math) and I’m confused so here’s the question
hjlf
I think it’s 5 hrs I’m not sur though but that was I think it is
6 0
3 years ago
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What is the graph of the function f(x)=x2+ 2x+ 3?
8090 [49]

Answer:


Step-by-step explanation:

The best way to do this is to let a graphing program do it. You could do it from a chart, but a graphing program can be useful for that as well. We'll make up a mini chart here

x        method                        y

-3      (-3)^2 + 2(-3) + 3 =       6

-2      (-2)^2 + 2(-2) + 3 =       3

-1       (-1)^2 + 2(-1)  + 3 =        2

0                                            3

1        (1)^2 + 2(1) + 3              6

2        (2^2) + 2(2) + 3           11

3        (3)^2  +2(3) + 3            18  

See the graph below.    

You can pick it out from the graphs you were given.                

8 0
4 years ago
The product of 1 1/2 and 2 is
Over [174]

Method 1:

Convert the mixed number to the improper fraction:

1\dfrac{1}{2}=\dfrac{1\cdot2+1}{2}=\dfrac{3}{2}

Make the product:

1\dfrac{1}{2}\cdot2=\dfrac{3}{2}\cdot2

<em>canceled 2</em>

=\dfrac{3}{\not2_1}\cdot\not2^1}=\boxed{3}

Method 2:

1\dfrac{1}{2}=1+\dfrac{1}{2}

1\dfrac{1}{2}\cdot2=\left(1+\dfrac{1}{2}\right)\cdot2

<em>use the distributive property a(b + c) = ab + ac</em>

(1)(2)+\left(\dfrac{1}{2}\right)(2)=2+1=\boxed{3}

7 0
3 years ago
HELP PLEASE SOON!
kirill115 [55]
1.\\\\\bold{v_1}=\langle2,4\rangle\qquad\bold{v_2}=\langle-1,5\rangle\\\\&#10;\bold{v_1}\circ\bold{v_2}=2\cdot(-1)+4\cdot5=-2+20=18\qquad [\text{C}]\\\\\\&#10;2.\\\\&#10;\bold{v_1}= \langle2, 5\rangle\qquad \bold{v_2}= \langle4, -3\rangle\\\\\\&#10;\bold{v_1}\circ\bold{v_2}=2\cdot4+5\cdot(-3)=8-15=-7\\\\||\bold{v_1}||=\sqrt{2^2+5^2}=\sqrt{4+25}=\sqrt{29}\\\\&#10;||\bold{v_2}||=\sqrt{4^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5\\\\\\&#10;

\cos\Theta=\dfrac{\bold{v_1}\circ\bold{v_2}}{||\bold{v_1}||\cdot||\bold{v_2}||}=\dfrac{-7}{5\cdot\sqrt{29}}\\\\\\&#10;\cos\Theta=-\dfrac{7}{5\sqrt{29}}\quad\implies\quad\Theta=\arccos\left(-\dfrac{7}{5\sqrt{29}}\right)\\\\\\\boxed{\Theta\approx105,1^\circ}
7 0
4 years ago
Read 2 more answers
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