3 years ago

URGENT Can someone please help me with this? Thank you. :)

Mathematics

1 answer:

vovangra [49]3 years ago

8 0

Answer:

Step-by-step explanation:

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What is the slope of the line that passes through the points El 1, 4) and F(2, 6)?

Alex787 [66]

Answer:

2/3

Step-by-step explanation:

(6 -4) / (2 - (-1)) =(2) / (2+1) = 2/3

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3 years ago

Liam spent $4.76 on a salad bowl, and $3.81 on a cup of coffee. He paid using TWO ten dollar bills. What was Liam's change?

quester [9]

Answer:

11.43

Step-by-step explanation:

You can start this by adding the total cost of the salad and the coffee.

4.76 + 3.81 = 8.57

Then you have the paid amount of 20 bucks because there was two tens.

20 - 8.57 = 11.43.

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3 years ago

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Simplify the expression 4 + 23−|−4| <br> A.16 B.4 C.8 D.12

Artyom0805 [142]

Answer: =23

Step-by-step explanation:

$4+23-\left|-4\right|$

$=4+23-4$

$=23$

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3 years ago

Gabrielle's age is two times mikhall's age. the sum of their ages is 54. what is mikhall's age?

Elodia [21]

Answer:

Step-by-step explanation:

54/2

27x2=54 and they said 2 times

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3 years ago

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Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

3 0

3 years ago

***URGENT*** Can someone please help me with this? Thank you. :)

URGENT Can someone please help me with this? Thank you. :)